Algebra 2 · N.RN.A.1

Higher roots

A number multiplied by itself four, five or six times gives a known result. Which number was it, and when is there one answer, two, or none?

You need

Cube roots and higher rootsMath 8

Unlocks

Rational exponentsAlgebra 2

Where you are 3 in Square and cube roots

Number9

Exponents and roots5

Square and cube roots3

Square roots Cube roots and higher roots

Higher roots

Why this exists

A square root undoes squaring and a cube root undoes cubing. Nothing stops at three. A quantity that is multiplied by the same factor every year for four years has grown by that factor used four times, and to recover the yearly factor from the four-year total you need a number whose fourth power is known. Equations such as x^5 = -243 and x^6 = 729 ask the same kind of question with five and six equal factors.

The cube roots lesson met these higher roots in passing. This lesson treats them in full, and one fact runs through all of it: whether the index is even or odd decides almost everything. It decides whether a negative number has a root, whether an equation has one solution or two, whether undoing a power gives back x or only |x|, and when the rules for products and quotients may be used at all.

Definition

Let n be a whole number, 2 or more. A number b is an n-th root of a number a when b used n times as a factor gives a:

b^n = a

The radical sign names one particular root, the principal n-th root, written \displaystyle \sqrt[n]{a}. The number n is the index and a is the radicand. For the square root the index 2 is left unwritten, so \displaystyle \sqrt{a} means \displaystyle \sqrt[2]{a}. Which root the sign names depends on whether n is odd or even.

Odd index (n = 3, 5, 7, \ldots). Every real number a has exactly one real n-th root, and \displaystyle \sqrt[n]{a} is that root. It has the same sign as a. For example \displaystyle \sqrt[5]{-243} = -3 because (-3)^5 = -243.

Even index (n = 2, 4, 6, \ldots). Only a radicand a \geq 0 has a real n-th root. A positive a has two of them, one positive and one negative, and \displaystyle \sqrt[n]{a} means the one that is not negative. A negative a has none, so for example \displaystyle \sqrt[6]{-64} is not a real number.

So in every case,

\begin{aligned} \sqrt[n]{a} = b \quad \text{means} \quad &b^n = a, \\ &\text{and } b \geq 0 \text{ if } n \text{ is even} \end{aligned}

A perfect n-th power is the n-th power of an integer, such as 2^4 = 16, 3^5 = 243 or 2^6 = 64. Its principal n-th root is an integer. A short table of powers is enough for most of this lesson:

b	b^4	b^5	b^6
2	16	32	64
3	81	243	729
4	256	1024	4096
5	625	3125	15\,625
10	10\,000	100\,000	1\,000\,000

Intuition

Everything about the index comes down to what a power does to a sign. A product of negative factors is positive when there is an even number of them, because they pair off, and negative when there is an odd number, because one is left over. So

\begin{aligned} (-b)^n &= b^n && n \text{ even} \\ (-b)^n &= -b^n && n \text{ odd} \end{aligned}

An even power wipes the sign out. Then b and -b have the same n-th power, and no power is ever negative. An odd power keeps the sign. Then positive numbers go to positive powers, negative numbers to negative powers, and no two numbers share a power.

Figure 1 shows this as two graphs. The graph of y = x^4 is shaped like a U: a level line above the bottom meets it twice, at a number and its opposite, and a level line below the x-axis misses it entirely. The graph of y = x^5 climbs from left to right without ever turning back, so every level line, above or below the axis, meets it exactly once. Every even power has the first shape and every odd power the second.

Figure 1An even power, like y = x^4, is never negative and gives each positive value twice. An odd power, like y = x^5, gives every value exactly once.

The same sign fact explains what happens when a root undoes a power. Take x = -2. Its fifth power is -32, and the fifth root of -32 is -2 again, so the fifth root gives back exactly what went in. Its sixth power is 64, the same as the sixth power of 2, and the sixth root of 64 is 2: the sign was lost in the power and the root cannot bring it back. Figure 2 graphs both. The odd case is the straight line y = x; the even case is the V shape y = |x|, the absolute value.

Figure 2The sixth root of x^6 is |x|, a V shape. The fifth root of x^5 is x itself, a straight line.

Results

The index parity rule. For a real number p, the equation x^n = p has, over the real numbers,

for odd n: exactly one solution, \displaystyle x = \sqrt[n]{p}, for every p;
for even n: two solutions \displaystyle x = \pm\sqrt[n]{p} when p > 0, one solution x = 0 when p = 0, and none when p < 0.

A root undoes its power. Whenever \displaystyle \sqrt[n]{a} is a real number,

\left(\sqrt[n]{a}\right)^n = a

Taken the other way round, for every real number a,

\begin{aligned} \sqrt[n]{a^n} &= a && n \text{ odd} \\ \sqrt[n]{a^n} &= |a| && n \text{ even} \end{aligned}

Negatives through an odd root. For odd n and every real a,

\sqrt[n]{-a} = -\sqrt[n]{a}

Products and quotients. When \displaystyle \sqrt[n]{a} and \displaystyle \sqrt[n]{b} are both real numbers (for odd n, any a and b; for even n, a \geq 0 and b \geq 0), and b \neq 0 in the quotient,

\begin{aligned} \sqrt[n]{ab} &= \sqrt[n]{a}\,\sqrt[n]{b} \\ \sqrt[n]{\frac{a}{b}} &= \frac{\sqrt[n]{a}}{\sqrt[n]{b}} \end{aligned}

A root of a root. For a \geq 0, and for every real a when m and n are both odd,

\sqrt[m]{\sqrt[n]{a}} = \sqrt[mn]{a}

In particular a fourth root is a square root taken twice, \displaystyle \sqrt[4]{a} = \sqrt{\sqrt{a}}.

Why it is true

The parity rule. Take two numbers with 0 \leq a < b. Multiplying the inequality a < b by itself n times, with every factor 0 or more, keeps its direction, so a^n < b^n. Bigger numbers that are not negative have bigger n-th powers.

For odd n the sign comes through, (-b)^n = -b^n, so the same order holds among negative numbers, and a negative number always has a smaller power than a positive one. Two different numbers never share an odd power, so x^n = p has at most one solution. That it has at least one, for every p, is what Figure 1 shows: as x runs from large negative to large positive values, x^n runs through every value with no jumps. That step is argued from the graph and not proved here; it belongs to later work on continuous functions.

For even n, write n = 2k. Then x^n = (x^k)^2 is a square, so it is never negative and x^n = p has no solution when p < 0. Among numbers 0 or more there is exactly one solution when p \geq 0, by the order fact above and the same graph argument. Its opposite is a second solution, because (-x)^n = x^n, and it is different from the first unless both are 0.

Even roots of even powers. Let n be even. Since an even power wipes out the sign, a^n = |a|^n. So |a| is a number whose n-th power is a^n, and |a| \geq 0. The principal root \displaystyle \sqrt[n]{a^n} is the only number with both properties, so it equals |a|. For odd n, a itself has n-th power a^n, and only one number does, so \displaystyle \sqrt[n]{a^n} = a.

Products, quotients and negatives. Raise the right side to the n-th power. Regrouping and reordering the factors (the associative and commutative properties) gives

\left(\sqrt[n]{a}\,\sqrt[n]{b}\right)^n = \left(\sqrt[n]{a}\right)^n\left(\sqrt[n]{b}\right)^n = ab

For odd n, only one number has n-th power ab, so \displaystyle \sqrt[n]{a}\,\sqrt[n]{b} = \sqrt[n]{ab}. For even n, up to two numbers have that power, but \displaystyle \sqrt[n]{a}\,\sqrt[n]{b} is a product of two numbers that are not negative, so it is the one that is not negative, which is \displaystyle \sqrt[n]{ab}. The quotient works the same way. For odd n, \displaystyle \left(-\sqrt[n]{a}\right)^n = -a shows \displaystyle \sqrt[n]{-a} = -\sqrt[n]{a}.

A root of a root. Powers of powers multiply, so

\begin{aligned} \left(\sqrt[m]{\sqrt[n]{a}}\right)^{mn} &= \left(\left(\sqrt[m]{\sqrt[n]{a}}\right)^m\right)^n \\ &= \left(\sqrt[n]{a}\right)^n = a \end{aligned}

When a \geq 0, the left side is a root of a root of a number that is not negative, so it is not negative itself, and the only such number with mn-th power a is \displaystyle \sqrt[mn]{a}. When m and n are both odd, mn is odd and only one number has mn-th power a.

Worked examples

Example 1

Roots of any index

Find \displaystyle \sqrt[5]{-243}, \displaystyle \sqrt[4]{1296}, \displaystyle \sqrt[7]{128}, \displaystyle \sqrt[6]{-64} and \displaystyle -\sqrt[6]{64}.

Solution

For \displaystyle \sqrt[5]{-243} the index 5 is odd, so the negative radicand has one root and it is negative. From the table, 3^5 = 243, so try -3: (-3)^5 = -243 because five negative factors leave one unpaired. So \displaystyle \sqrt[5]{-243} = -3.

The number 1296 is not in the table, so use the root of a root: a fourth root is a square root taken twice.

\sqrt[4]{1296} = \sqrt{\sqrt{1296}} = \sqrt{36} = 6

Check: 6^4 = 36 \times 36 = 1296. The answer -6 also has fourth power 1296, but the sign names the root that is not negative.

For \displaystyle \sqrt[7]{128}, the powers of 2 run 2, 4, 8, 16, 32, 64, 128, and 128 is the seventh. So \displaystyle \sqrt[7]{128} = 2.

For \displaystyle \sqrt[6]{-64} the index is even and the radicand negative. A sixth power is a square, b^6 = (b^3)^2, and is never negative, so \displaystyle \sqrt[6]{-64} is not a real number.

In \displaystyle -\sqrt[6]{64} the minus sign is outside the root. Take the root first: 2^6 = 64, so \displaystyle \sqrt[6]{64} = 2, and then \displaystyle -\sqrt[6]{64} = -2. With an odd index a minus inside or in front gives the same thing; with an even index one is a real number and the other is not.

Example 2

Fractions and decimals

Find \displaystyle \sqrt[4]{0.0016} and \displaystyle \sqrt[5]{-\frac{32}{243}}.

Solution

Write the decimal as a fraction: \displaystyle 0.0016 = \frac{16}{10\,000}. Both parts are fourth powers, 16 = 2^4 and 10\,000 = 10^4, so by the quotient rule

\sqrt[4]{0.0016} = \frac{\sqrt[4]{16}}{\sqrt[4]{10\,000}} = \frac{2}{10} = 0.2

Check: 0.2^2 = 0.04 and 0.04^2 = 0.0016. The index is even, so the quotient rule needs both parts not negative, which they are.

For \displaystyle \sqrt[5]{-\frac{32}{243}} the index is odd, so the negative comes straight out of the root:

\sqrt[5]{-\frac{32}{243}} = -\frac{\sqrt[5]{32}}{\sqrt[5]{243}} = -\frac{2}{3}

Check: \displaystyle \left(-\frac{2}{3}\right)^5 = -\frac{2^5}{3^5} = -\frac{32}{243}.

Example 3

Equations, one index at a time

Solve over the real numbers: x^6 = 729, 2x^5 = -486, x^4 + 40 = 15 and (x - 2)^4 = 1296.

Solution

For x^6 = 729 the index is even and 729 is positive, so the parity rule promises two solutions. 3^6 = 729, and (-3)^6 = 729 as well, so x = 3 or x = -3.

For 2x^5 = -486, divide both sides by 2 to get x^5 = -243. The index is odd, so there is exactly one solution, \displaystyle x = \sqrt[5]{-243} = -3.

For x^4 + 40 = 15, subtract 40 from both sides: x^4 = -25. A fourth power is never negative, so this equation has no real solution. There is nothing to compute; the sign settles it.

For (x - 2)^4 = 1296, the quantity x - 2 has fourth power 1296, so it is one of the two fourth roots, 6 or -6:

x - 2 = 6 \quad \text{or} \quad x - 2 = -6

Adding 2 to both sides of each gives x = 8 or x = -4. Check: (8 - 2)^4 = 6^4 = 1296 and (-4 - 2)^4 = (-6)^4 = 1296. Taking only the principal root here would have lost x = -4.

Example 4

Even roots of powers

Simplify, for every real x and y: \displaystyle \sqrt[6]{x^6}, \displaystyle \sqrt[5]{x^5}, \displaystyle \sqrt[4]{16y^4}, \displaystyle \sqrt[4]{x^8} and \displaystyle \sqrt[4]{x^{12}}.

Solution

The index 6 is even, so \displaystyle \sqrt[6]{x^6} = |x|. Trying x = -2 shows why the absolute value is needed: \displaystyle \sqrt[6]{(-2)^6} = \sqrt[6]{64} = 2, which is |-2|, not -2.

The index 5 is odd, so \displaystyle \sqrt[5]{x^5} = x for every x. For x = -2, (-2)^5 = -32 and its fifth root is -2.

For \displaystyle \sqrt[4]{16y^4} both factors are not negative, so the product rule splits it: \displaystyle \sqrt[4]{16}\,\sqrt[4]{y^4} = 2|y|.

For \displaystyle \sqrt[4]{x^8}, write x^8 = (x^2)^4. So \displaystyle \sqrt[4]{x^8} is the fourth root of the fourth power of x^2, which is |x^2|. But x^2 is never negative, so |x^2| = x^2 and no absolute value sign is needed:

\sqrt[4]{x^8} = x^2

For \displaystyle \sqrt[4]{x^{12}}, write x^{12} = (x^3)^4. Now the base x^3 can be negative, so the absolute value must stay: \displaystyle \sqrt[4]{x^{12}} = |x^3|, which can also be written |x|^3. For x = -1 the left side is \displaystyle \sqrt[4]{1} = 1, while x^3 = -1; only |x^3| = 1 is right.

The question to ask each time is whether the base of the power, once the radicand is written as a base to the n-th power, can be negative.

Example 5

Multiplying and dividing roots

Find \displaystyle \sqrt[4]{8} \cdot \sqrt[4]{2}, \displaystyle \sqrt[5]{9} \cdot \sqrt[5]{-27} and \displaystyle \frac{\sqrt[4]{48}}{\sqrt[4]{3}}. Then write \displaystyle \sqrt[4]{48} with the smallest possible radicand.

Solution

Neither \displaystyle \sqrt[4]{8} nor \displaystyle \sqrt[4]{2} is an integer, but both radicands are positive, so the product rule joins them into one root:

\sqrt[4]{8} \cdot \sqrt[4]{2} = \sqrt[4]{16} = 2

For \displaystyle \sqrt[5]{9} \cdot \sqrt[5]{-27} the index is odd, so the product rule holds even with a negative factor:

\sqrt[5]{9} \cdot \sqrt[5]{-27} = \sqrt[5]{-243} = -3

For the quotient, both radicands are positive and the bottom is not zero:

\frac{\sqrt[4]{48}}{\sqrt[4]{3}} = \sqrt[4]{\frac{48}{3}} = \sqrt[4]{16} = 2

Read the same way in reverse, 48 = 16 \times 3 and 16 is a fourth power, so \displaystyle \sqrt[4]{48} = \sqrt[4]{16}\,\sqrt[4]{3} = 2\sqrt[4]{3}. No fourth power other than 1 divides 3, so 3 is the smallest radicand. Pulling out perfect squares from a square root the same way is the subject of simplifying radicals; a higher root works just like this.

Example 6

Where the rules stop

Compare \displaystyle \sqrt[4]{(-2)(-8)} with \displaystyle \sqrt[4]{-2} \cdot \sqrt[4]{-8}, and \displaystyle \sqrt[5]{(-2)(-16)} with \displaystyle \sqrt[5]{-2} \cdot \sqrt[5]{-16}. Then decide whether \displaystyle \sqrt[4]{16 + 16} equals \displaystyle \sqrt[4]{16} + \sqrt[4]{16}.

Solution

Inside the first root, (-2)(-8) = 16, so \displaystyle \sqrt[4]{(-2)(-8)} = \sqrt[4]{16} = 2. But \displaystyle \sqrt[4]{-2} and \displaystyle \sqrt[4]{-8} are even roots of negative numbers and are not real, so the product \displaystyle \sqrt[4]{-2} \cdot \sqrt[4]{-8} is not a real number at all. The product rule fails here because its condition fails: for an even index both radicands must be 0 or more.

With the odd index 5 everything is real. (-2)(-16) = 32, so \displaystyle \sqrt[5]{(-2)(-16)} = \sqrt[5]{32} = 2. On the other side, \displaystyle \sqrt[5]{-2} and \displaystyle \sqrt[5]{-16} are both real and negative, and the rule says their product is \displaystyle \sqrt[5]{32} = 2, a positive number, as a product of two negatives should be.

For the sum, \displaystyle \sqrt[4]{16} + \sqrt[4]{16} = 2 + 2 = 4. The left side is \displaystyle \sqrt[4]{32}, and since 2^4 = 16 < 32 < 81 = 3^4, it lies between 2 and 3. So the two are not equal. A root splits over a product, never over a sum: the proof raised a product to the n-th power, and a sum raised to a power has extra terms.

Example 7

A yearly growth factor

Kaelan counts the birds in a colony. Over four years the count went from 256 to 1296, and each year it was multiplied by the same factor. What was the factor, and what were the counts in between?

Solution

Call the yearly factor r. After four years the count has been multiplied by r four times, so

256r^4 = 1296

Dividing both sides by 256 gives \displaystyle r^4 = \frac{1296}{256}. The index is even, so there are two real fourth roots, and the quotient rule gives the positive one:

r = \sqrt[4]{\frac{1296}{256}} = \frac{\sqrt[4]{1296}}{\sqrt[4]{256}} = \frac{6}{4} = \frac{3}{2}

The other root, \displaystyle -\frac{3}{2}, would make the count negative every other year, which a count of birds cannot be, so the factor is \displaystyle r = \frac{3}{2}: the colony grew by half each year. The counts run

256, \quad 384, \quad 576, \quad 864, \quad 1296

each one \displaystyle \frac{3}{2} times the one before. Check: \displaystyle 864 \times \frac{3}{2} = 1296.

Boundaries

The index is a whole number, 2 or more. A first root would just be the number itself. A root is not written with a fraction as its index; fractions appear instead as exponents, in rational exponents, where \displaystyle \sqrt[n]{a} is written a^{1/n}.

Even roots stop at zero. For even n, \displaystyle \sqrt[n]{a} is a real number only when a \geq 0. The complex numbers supply roots for negative radicands, and also extra roots for positive ones (in the complex numbers x^4 = 16 has four solutions), but in this lesson the radical sign always names a real number.

The radical sign names one root. For even n it is the root that is not negative, so \displaystyle \sqrt[4]{1296} = 6, never -6. Solving x^n = p for even n needs both \displaystyle \pm\sqrt[n]{p}; the sign alone gives one of them.

An even root of an even power is an absolute value. \displaystyle \sqrt[n]{a^n} = |a| for even n, and equals a only when a \geq 0. The absolute value may be dropped only when the base is known not to be negative, as with \displaystyle \sqrt[4]{x^8} = x^2.

The product and quotient rules need real roots on the right. For even n both radicands must be 0 or more. Joining \displaystyle \sqrt[4]{-2} \cdot \sqrt[4]{-8} into \displaystyle \sqrt[4]{16} turns something that is not a real number into 2. For odd n there is no restriction, apart from a zero denominator.

No rule for sums or differences. \displaystyle \sqrt[n]{a + b} is not \displaystyle \sqrt[n]{a} + \sqrt[n]{b}, for any index.

A root of a root needs care with signs. \displaystyle \sqrt[m]{\sqrt[n]{a}} = \sqrt[mn]{a} holds for a \geq 0, and for negative a only when both indexes are odd.

Where this goes next

Roots of any index are the doorway to rational exponents. Writing \displaystyle \sqrt[n]{a} as a^{1/n} turns the rules of this lesson into the familiar rules of exponents: the product rule becomes (ab)^{1/n} = a^{1/n} b^{1/n}, and the root of a root becomes (a^{1/n})^{1/m} = a^{1/(mn)}. The parity conditions travel with them, which is why a rational power of a negative number needs the same care as an even root.

The product rule read in reverse is how a root is simplified, by pulling out the largest perfect n-th power, as \displaystyle \sqrt[4]{48} = 2\sqrt[4]{3} did above; for square roots that is the work of simplifying radicals. The parity rule reappears whenever an equation has a power in it: x^n = p is the simplest polynomial equation, and its one, two or no real solutions are the first sign of how the number of solutions of a polynomial depends on its degree. And the growth factor of the last example, a fourth root of a total change, is exactly how a steady yearly rate is found from the start and end of any period of repeated multiplication.

Exercises

Roots of any index

Find \displaystyle \sqrt[6]{729}.
Find \displaystyle \sqrt[5]{-100000}.
Find \displaystyle \sqrt[8]{256}.
Find \displaystyle \sqrt[4]{-1296}, or say why it is not a real number.
Find \displaystyle -\sqrt[4]{2401}.

Fractions and decimals

Find \displaystyle \sqrt[4]{\frac{256}{2401}}.
Find \displaystyle \sqrt[5]{-0.00032}.
Find \displaystyle \sqrt[6]{0.000001}.

Solving equations

Solve each equation over the real numbers.

x^4 = 2401
\displaystyle x^5 = -\frac{1}{243}
3x^6 = 192
x^4 + 30 = 5
(2x - 1)^4 = 16
(x + 3)^5 = -1

Even roots of powers

Simplify each expression, where every letter stands for any real number.

\displaystyle \sqrt[8]{x^8}
\displaystyle \sqrt[6]{64y^6}
\displaystyle \sqrt[6]{x^{12}}
\displaystyle \sqrt[6]{x^{18}}
\displaystyle \sqrt[5]{-243a^{10}}

Multiplying and dividing roots

Find \displaystyle \sqrt[6]{16} \cdot \sqrt[6]{4}.
Find \displaystyle \frac{\sqrt[4]{160}}{\sqrt[4]{10}}.
Find \displaystyle \sqrt[5]{-4} \cdot \sqrt[5]{-8}, and say why the product rule may be used.

Reasoning

Sundari writes \displaystyle \sqrt[6]{-2} \cdot \sqrt[6]{-32} = \sqrt[6]{64} = 2. What is wrong with this?
Explain why \displaystyle \sqrt[n]{a^n} = |a| for every even index n and every real a. For which real numbers a is \displaystyle \sqrt[n]{a^n} = a when n is even?
Show that \displaystyle \sqrt{\sqrt[3]{a}} = \sqrt[6]{a} for every a \geq 0, and check it for a = 64. What happens to each side when a < 0?
Explain why \displaystyle \sqrt[n]{-a} = -\sqrt[n]{a} for every odd index n and every real a, and use it with Example 1 to find \displaystyle \sqrt[7]{-128}.

Answers

3, since 3^6 = 729 and 3 is positive.
-10. The index is odd, and (-10)^5 = -100\,000.
2, since 2^8 = 256 and 2 is positive.
Not a real number. The index 4 is even and the radicand is negative, and a fourth power is a square, so it is never negative.
-7. First \displaystyle \sqrt[4]{2401} = 7, since 7^2 = 49 and 49^2 = 2401; the sign in front makes it -7.
\displaystyle \frac{4}{7}, since 4^4 = 256 and 7^4 = 2401, by the quotient rule.
-0.2. \displaystyle 0.00032 = \frac{32}{100\,000}, with 2^5 = 32 and 10^5 = 100\,000, so the root is \displaystyle -\frac{2}{10}. The odd index lets the negative come straight through.
0.1. \displaystyle 0.000001 = \frac{1}{10^6}, whose principal sixth root is \displaystyle \frac{1}{10}.
x = 7 or x = -7. The index is even and 2401 > 0, so there are two roots, \displaystyle \pm\sqrt[4]{2401}.
\displaystyle x = -\frac{1}{3}, the one real fifth root, since 3^5 = 243 and an odd power keeps the sign.
x = 2 or x = -2. Divide by 3: x^6 = 64, and 2^6 = (-2)^6 = 64.
No real solution. Subtracting 30 gives x^4 = -25, and a fourth power is never negative.
\displaystyle x = \frac{3}{2} or \displaystyle x = -\frac{1}{2}. The base 2x - 1 is 2 or -2, so 2x = 3 or 2x = -1. Check: (3 - 1)^4 = 16 and (-1 - 1)^4 = 16.
x = -4. The odd index gives one root: \displaystyle x + 3 = \sqrt[5]{-1} = -1.
|x|. The index is even, so the root of the eighth power gives back x without its sign.
2|y|. \displaystyle \sqrt[6]{64} = 2 and \displaystyle \sqrt[6]{y^6} = |y|, and both radicands are not negative.
x^2. x^{12} = (x^2)^6, and x^2 is never negative, so no absolute value is needed.
|x^3|, which is also |x|^3. x^{18} = (x^3)^6, and x^3 is negative when x is.
-3a^2. The index is odd, so no absolute value is needed: (-3a^2)^5 = -243a^{10}.
2. The radicands are positive, so \displaystyle \sqrt[6]{16} \cdot \sqrt[6]{4} = \sqrt[6]{64} = 2.
2. \displaystyle \frac{160}{10} = 16 and \displaystyle \sqrt[4]{16} = 2.
2. The index 5 is odd, so both roots are real and the rule holds for any signs: \displaystyle \sqrt[5]{(-4)(-8)} = \sqrt[5]{32} = 2.
The index 6 is even, so \displaystyle \sqrt[6]{-2} and \displaystyle \sqrt[6]{-32} are not real numbers and the left side is not a real number. The product rule for an even index needs both radicands to be 0 or more, so it cannot be used here; joining the roots turned something undefined into 2.
An even power wipes out the sign, (-a)^n = a^n, so a^n = |a|^n. Then |a| is a number that is not negative and has n-th power a^n, and the principal root is the only such number, so \displaystyle \sqrt[n]{a^n} = |a|. This equals a exactly when |a| = a, that is, when a \geq 0.
For a \geq 0, square \displaystyle \sqrt{\sqrt[3]{a}} to get \displaystyle \sqrt[3]{a}, then cube that to get a. So \displaystyle \left(\sqrt{\sqrt[3]{a}}\right)^6 = a, and \displaystyle \sqrt{\sqrt[3]{a}} is not negative, so it is the principal sixth root \displaystyle \sqrt[6]{a}. For a = 64: \displaystyle \sqrt{\sqrt[3]{64}} = \sqrt{4} = 2 and \displaystyle \sqrt[6]{64} = 2. For a < 0, \displaystyle \sqrt[3]{a} is negative, so its square root is not real, and \displaystyle \sqrt[6]{a} is not real either.
For odd n, \displaystyle \left(-\sqrt[n]{a}\right)^n = (-1)^n\left(\sqrt[n]{a}\right)^n = -a, and only one real number has n-th power -a, so it is \displaystyle \sqrt[n]{-a}. So \displaystyle \sqrt[7]{-128} = -\sqrt[7]{128} = -2.