Math 8 · 8.EE.A.2 · N.RN.A.1

Cube roots and higher roots

A cube-shaped box holds a known amount. How long is each edge, and which number, used three times as a factor, gives that amount back?

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Square roots

Cube roots and higher roots

Higher roots

Why this exists

A square root runs squaring backwards. Told that a square has area 49, you ask which number times itself gives 49, and the side is 7.

A cube asks the same question one dimension up. A cube-shaped water tank or a box packed with dice has a volume, and the volume is the edge used three times as a factor, V = s \cdot s \cdot s = s^3 (the volume of a prism with every edge equal). Going from edge to volume is multiplication. Going from volume back to edge needs a new operation, one that undoes cubing. That operation is the cube root.

It also settles equations. To solve x^3 = -512 you need a number whose cube is -512, and a square root is no help: no number squared is negative. A cube root is, because a negative number cubed stays negative. Cubing and squaring differ in a second way: two different numbers can have the same square, but never the same cube. That is why x^2 = 64 has two solutions and x^3 = 64 has exactly one.

Definition

A number b is a cube root of a number a when b cubed is a:

b^3 = a

Every real number has exactly one real cube root, and it is written \displaystyle \sqrt[3]{a}. The small 3 is the index; it says how many equal factors are being undone. So

\sqrt[3]{a} = b \quad \text{means} \quad b^3 = a

For example \displaystyle \sqrt[3]{64} = 4 because 4^3 = 64, and \displaystyle \sqrt[3]{-64} = -4 because (-4)^3 = (-4)(-4)(-4) = -64.

A perfect cube is the cube of an integer: 0, 1, 8, 27, 64 and so on, and also -1, -8, -27 and so on. The cube root of a perfect cube is an integer. Knowing the first ten cubes by sight makes the cube roots of perfect cubes quick to read off, and each negative cube is the opposite of the one shown:

n	n^3	n	n^3
1	1	6	216
2	8	7	343
3	27	8	512
4	64	9	729
5	125	10	1000

Higher roots work the same way: b is an n-th root of a when b^n = a. For odd n, \displaystyle \sqrt[n]{a} is the one real n-th root, with the same sign as a. For even n, \displaystyle \sqrt[n]{a} is the root that is not negative, as with \displaystyle \sqrt{a}, and a negative a has none. This lesson only looks ahead to them; higher roots treats them in full.

Intuition

Picture a cube-shaped box packed with dice, as in Figure 1. One layer holds 4 \times 4 = 16 dice, and four layers hold 64. The cube root runs this in reverse: given 64 dice to build a solid cube, find the number that, used three times as a factor, makes 64.

Figure 1Four dice along each edge fill the box with 4^3 = 64 dice, so \displaystyle \sqrt[3]{64} = 4.

Signs work differently from squares. Multiplying two negatives gives a positive, so a square is never negative. Multiplying three negatives gives a positive from the first two and then a negative from the third, so the cube of a negative number is negative. Positive numbers cube to positive numbers, negatives to negatives, and 0 to 0. That is why a negative number can have a cube root at all.

Figure 2 shows the second difference. The graph of y = x^2 comes down and goes back up, so a level line above the bottom crosses it twice, at x = 2 and x = -2 for the line y = 4, and a line below it misses entirely. The graph of y = x^3 climbs the whole way from left to right and never turns back, so every level line, high or low, crosses it exactly once.

Figure 2A level line meets y = x^2 twice, once (at 0) or not at all. It meets y = x^3 exactly once, wherever it is drawn.

Results

Cube root undoes cube. For every real number a,

\left(\sqrt[3]{a}\right)^3 = a \qquad \text{and} \qquad \sqrt[3]{a^3} = a

Negatives come straight through.

\sqrt[3]{-a} = -\sqrt[3]{a}

Products and quotients split. For real numbers a and b, with b \neq 0 in the quotient,

\sqrt[3]{ab} = \sqrt[3]{a}\,\sqrt[3]{b} \qquad \sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}}

Solving a cube equation. For every real number p, the equation x^3 = p has exactly one real solution,

x = \sqrt[3]{p}

Edge from volume. A cube with volume V has edge

s = \sqrt[3]{V}

Higher roots. For odd n, x^n = p has exactly one real solution, \displaystyle \sqrt[n]{p}. For even n, x^n = p has two real solutions \displaystyle \pm\sqrt[n]{p} when p > 0, one solution 0 when p = 0, and none when p < 0.

Why it is true

Two things are needed: that x^3 = p has at most one solution, which is proved here, and that it has at least one, which is argued from the graph and not fully proved.

Different numbers have different cubes. Take two real numbers a < b.

If both are 0 or more, multiply the inequality a < b by itself three times. Each factor is at least 0, and multiplying inequalities between numbers that are not negative keeps the direction, so a^3 < b^3.

If both are 0 or less, then -b < -a and both of those are 0 or more, so by the first case (-b)^3 < (-a)^3. That says -b^3 < -a^3, and multiplying by -1 turns it round: a^3 < b^3.

If a < 0 < b, then a^3 is negative and b^3 is positive, so again a^3 < b^3.

In every case the larger number has the larger cube, so x^3 = p has at most one solution. This is exactly where squaring is different: -8 < 8 but (-8)^2 = 8^2, because squaring wipes out the sign.

Every number is reached (not proved here). As x slides from large negative numbers to large positive ones, x^3 slides from large negative values to large positive ones with no jumps, so the graph in Figure 2 meets every level line. For a perfect cube the solution can simply be shown, as (-8)^3 = -512 shows. For a number like 300 the graph is the honest argument at this stage; the proof that no value is skipped belongs to later work on continuous functions.

The rules for products, quotients and negatives all come from that uniqueness. To show \displaystyle \sqrt[3]{ab} = \sqrt[3]{a}\,\sqrt[3]{b}, cube the right side. The factors can be regrouped and reordered (the associative and commutative properties), so

\left(\sqrt[3]{a}\,\sqrt[3]{b}\right)^3 = \left(\sqrt[3]{a}\right)^3\left(\sqrt[3]{b}\right)^3 = ab

The right side is a number whose cube is ab, and only one number has that cube, so it is \displaystyle \sqrt[3]{ab}. The quotient works the same way, and \displaystyle \sqrt[3]{-a} = -\sqrt[3]{a} follows because \displaystyle \left(-\sqrt[3]{a}\right)^3 = -a.

Even roots of negatives. An even power is a square, for instance b^4 = (b^2)^2, and a square is never negative, so no negative number has a real even root.

Worked examples

Example 1

Cube roots of perfect cubes

Find \displaystyle \sqrt[3]{343}, \displaystyle \sqrt[3]{-1000} and \displaystyle -\sqrt[3]{216}.

Solution

For \displaystyle \sqrt[3]{343}, look for the number whose cube is 343. From the table, 7^3 = 343, so

\sqrt[3]{343} = 7

For \displaystyle \sqrt[3]{-1000}, the number inside is negative, and a negative cube comes from a negative number. Since 10^3 = 1000, try -10: (-10)^3 = -1000. So \displaystyle \sqrt[3]{-1000} = -10.

In \displaystyle -\sqrt[3]{216} the minus sign sits outside the root, so take the root first and then the opposite. 6^3 = 216 gives \displaystyle \sqrt[3]{216} = 6, so \displaystyle -\sqrt[3]{216} = -6.

For a cube root a minus sign inside or in front gives the same result, since \displaystyle \sqrt[3]{-216} = -\sqrt[3]{216}.

Example 2

Fractions and decimals

Find \displaystyle \sqrt[3]{\frac{64}{27}}, \displaystyle \sqrt[3]{0.027} and \displaystyle \sqrt[3]{-\frac{1}{64}}.

Solution

A cube root of a quotient is the quotient of the cube roots. 64 = 4^3 and 27 = 3^3, so

\sqrt[3]{\frac{64}{27}} = \frac{\sqrt[3]{64}}{\sqrt[3]{27}} = \frac{4}{3}

Check: \displaystyle \left(\frac{4}{3}\right)^3 = \frac{4 \cdot 4 \cdot 4}{3 \cdot 3 \cdot 3} = \frac{64}{27}.

For the decimal, write it as a fraction first. \displaystyle 0.027 = \frac{27}{1000}, and both parts are perfect cubes:

\sqrt[3]{0.027} = \sqrt[3]{\frac{27}{1000}} = \frac{3}{10} = 0.3

Check: 0.3 \times 0.3 = 0.09 and 0.09 \times 0.3 = 0.027.

For \displaystyle \sqrt[3]{-\frac{1}{64}}, the negative comes straight through, and \displaystyle \sqrt[3]{1} = 1, \displaystyle \sqrt[3]{64} = 4:

\sqrt[3]{-\frac{1}{64}} = -\frac{1}{4}

Example 3

Solving a cube equation

Solve x^3 = -512, and then solve 4x^3 + 9 = 117.

Solution

For x^3 = -512 there is exactly one number whose cube is -512, the cube root:

x = \sqrt[3]{-512} = -8

Check: (-8)^3 = -512. No second answer is hiding, because two different numbers never have the same cube. Compare x^2 = 64, which has the two solutions 8 and -8, and x^2 = -64, which has none.

For 4x^3 + 9 = 117, get x^3 alone first, by the same moves as any equation: subtract 9 from both sides, then divide both sides by 4.

\begin{aligned} 4x^3 + 9 &= 117 \\ 4x^3 &= 108 \\ x^3 &= 27 \end{aligned}

Now take the cube root of both sides: \displaystyle x = \sqrt[3]{27} = 3.

Check: 4(3)^3 + 9 = 4(27) + 9 = 108 + 9 = 117.

Example 4

The edge of a cube-shaped tank

Matthias has a cube-shaped water tank that holds 1728 liters when full. One liter is 1000\ \text{cm}^3. How long is each inside edge? How deep is the water when the tank holds 432 liters?

Solution

Convert the volume to cubic centimeters so the edge comes out in centimeters:

V = 1728 \times 1000 = 1\,728\,000\ \text{cm}^3

The edge s satisfies s^3 = V, so \displaystyle s = \sqrt[3]{1\,728\,000}. Split it as a product: 1\,728\,000 = 1728 \times 1000, and \displaystyle \sqrt[3]{1000} = 10. For 1728, 12^2 = 144 and 144 \times 12 = 1728, so \displaystyle \sqrt[3]{1728} = 12. Then

s = 12 \times 10 = 120\ \text{cm}

For the depth, the water fills a box with the tank’s square floor. Its volume is the floor area times the depth, so the depth is the volume divided by the floor area:

\text{depth} = \frac{432\,000}{120 \times 120} = \frac{432\,000}{14\,400} = 30\ \text{cm}

Example 5

An edge that is not a whole number

Ravindra wants a cube-shaped box that holds exactly 300\ \text{cm}^3. Between which two whole numbers of centimeters does its edge lie? Narrow it to one decimal place.

Solution

The edge is \displaystyle \sqrt[3]{300}. Since 300 is not a perfect cube, this is the exact answer, and any decimal is only an approximation.

Bigger numbers have bigger cubes, so place 300 between two neighboring cubes:

6^3 = 216 < 300 < 343 = 7^3

The edge is between 6 and 7 cm, closer to 7, since 300 is much nearer to 343 than to 216. To narrow it, cube numbers with one decimal place near the top of that range:

6.6^3 = 287.496 \qquad 6.7^3 = 300.763

The first is below 300 and the second just above, so \displaystyle 6.6 < \sqrt[3]{300} < 6.7, and the edge is very close to 6.7 cm. A calculator gives \displaystyle \sqrt[3]{300} \approx 6.694, inside that range. The same squeezing is used for square roots in approximating irrational numbers.

Example 6

Fourth and fifth roots

Find \displaystyle \sqrt[4]{81}, \displaystyle \sqrt[4]{-16} and \displaystyle \sqrt[5]{-32}. Then solve x^4 = 256 over the real numbers.

Solution

3^4 = 3 \cdot 3 \cdot 3 \cdot 3 = 81, and 3 is positive, so \displaystyle \sqrt[4]{81} = 3.

For \displaystyle \sqrt[4]{-16}, the index is even, and a fourth power is a square, b^4 = (b^2)^2, so it is never negative. \displaystyle \sqrt[4]{-16} is not a real number.

For \displaystyle \sqrt[5]{-32}, the index is odd, so a negative number has a negative root, as with cube roots. (-2)^5 = -32, so \displaystyle \sqrt[5]{-32} = -2.

For x^4 = 256: 4^4 = 256, and (-4)^4 = 256 as well, because an even number of negative factors gives a positive product. So there are two solutions:

x = 4 \quad \text{or} \quad x = -4

An odd index would give one solution, as x^3 = p does.

Boundaries

One real cube root, not three. In the real numbers, x^3 = 8 has only the solution 2. Among the complex numbers it has two more, and \displaystyle \sqrt[3]{8} in this lesson always means the real one.

Even roots stop at zero. \displaystyle \sqrt{a}, \displaystyle \sqrt[4]{a} and every even root are real only for a \geq 0, and the symbol always means the root that is not negative. Odd roots have no such limit.

Undoing a power works cleanly only for odd powers. \displaystyle \sqrt[3]{a^3} = a for every a. For an even index, \displaystyle \sqrt[4]{a^4} = |a|, the absolute value: the fourth power erases the sign and the root cannot bring it back.

Roots split over products and quotients, not over sums. \displaystyle \sqrt[3]{ab} = \sqrt[3]{a}\,\sqrt[3]{b}, but \displaystyle \sqrt[3]{8 + 27} is \displaystyle \sqrt[3]{35}, which is not 2 + 3. The proof above cubed a product; a cubed sum has extra terms.

A decimal for a cube root is usually approximate. The cube root of a fraction in lowest terms is itself a fraction only when its top and bottom are both perfect cubes (an integer counts as a fraction over 1). Otherwise \displaystyle \sqrt[3]{p} is the exact answer and 6.694 is a rounded stand-in.

A length must be positive. The cube root of a negative number is a fine number, but a cube with volume -27 does not exist. In a measurement problem the volume is positive, so the edge is too.

Where this goes next

The next step is higher roots in full: the rules for any index, and simplifying a root like \displaystyle \sqrt[3]{54} = 3\sqrt[3]{2} by pulling out a perfect cube, which is the cube-root side of simplifying radicals. Later still, roots are written as powers with unit fraction exponents. The graph of \displaystyle y = \sqrt[3]{x}, Figure 2’s cubic turned on its side, is studied in cube root functions, and equations like x^3 = p grow into cubic and higher degree functions, where several real solutions can appear.

Most cube roots are not fractions at all: \displaystyle \sqrt[3]{300} is irrational, like \displaystyle \sqrt{2}. That is a fact stated here, not shown; the prime factor argument in irrational square roots carries over to cube roots with threes in place of twos. On the geometry side, the edge-from-volume move pairs with scale factors, where volumes scale by the cube of the factor and a cube root recovers the factor.

Exercises

Cube roots of perfect cubes

Find \displaystyle \sqrt[3]{64}.
Find \displaystyle \sqrt[3]{-27}.
Find \displaystyle \sqrt[3]{1331}.
Find \displaystyle -\sqrt[3]{-8}.
Find \displaystyle \sqrt[3]{8000}.

Fractions and decimals

Find \displaystyle \sqrt[3]{\frac{1}{216}}.
Find \displaystyle \sqrt[3]{-\frac{729}{1000}}, as a fraction and as a decimal.
Find \displaystyle \sqrt[3]{0.001}.
Find \displaystyle \sqrt[3]{0.216}.

Solving equations

Solve each equation over the real numbers.

x^3 = 729
\displaystyle x^3 = -\frac{8}{27}
5x^3 = -40
x^3 + 20 = -44
(x - 1)^3 = 27
Solve each of x^2 = 1, x^3 = 1, x^2 = -1 and x^3 = -1, and say how many real solutions each has.

Edges and volumes

Nkechi stacks 512 cube-shaped magnets, each with edge 1 cm, into one solid cube. How many magnets run along each edge, and how many show on the top face?
Quinlan-Jo packs a cube-shaped box exactly with 1000 dice, each a cube with edge 1.6 cm. Find the box’s inside edge in two ways: by counting dice along an edge, and from the total volume.
Oyelaran’s cube-shaped planter holds 64 liters of soil when full. One liter is 1000\ \text{cm}^3. How long is each inside edge?
A cube-shaped box holds 500\ \text{cm}^3. Between which two whole numbers of centimeters is its edge? Then decide whether the edge is more or less than 7.9 cm.
A cube-shaped box holds 3375\ \text{cm}^3. How much card covers its six faces?

Fourth and fifth roots

Find each value, or say why it is not a real number: \displaystyle \sqrt[4]{625}, \displaystyle \sqrt[5]{243}, \displaystyle \sqrt[4]{-81}, \displaystyle \sqrt[5]{-1}.
Solve over the real numbers: \displaystyle x^4 = \frac{1}{16}, x^5 = 32, x^4 = -1.

Reasoning

Explain why x^3 = p has exactly one real solution for every real number p, while x^2 = p can have two, one or none.
Is \displaystyle \sqrt[3]{8 + 27} equal to \displaystyle \sqrt[3]{8} + \sqrt[3]{27}? Work out the right side, place the left side between two whole numbers, and say what this shows.
Pernille says that \displaystyle \sqrt[4]{x^4} = x for every real x, just as \displaystyle \sqrt[3]{x^3} = x. Is she right? Explain, with an example.

Answers

4, since 4^3 = 64.
-3, since (-3)^3 = -27.
11, since 11^2 = 121 and 121 \times 11 = 1331.
2. First \displaystyle \sqrt[3]{-8} = -2, because (-2)^3 = -8; the sign outside makes it -(-2) = 2.
20. 8000 = 8 \times 1000, so \displaystyle \sqrt[3]{8000} = 2 \times 10 = 20. Check: 20^3 = 8000.
\displaystyle \frac{1}{6}, since \displaystyle \sqrt[3]{1} = 1 and \displaystyle \sqrt[3]{216} = 6.
\displaystyle -\frac{9}{10} = -0.9. The negative comes through, \displaystyle \sqrt[3]{729} = 9 and \displaystyle \sqrt[3]{1000} = 10.
0.1. \displaystyle 0.001 = \frac{1}{1000}, whose cube root is \displaystyle \frac{1}{10}.
0.6. \displaystyle 0.216 = \frac{216}{1000}, whose cube root is \displaystyle \frac{6}{10}. Check: 0.6^3 = 0.216.
x = 9, since 9^3 = 729.
\displaystyle x = -\frac{2}{3}, since \displaystyle \left(-\frac{2}{3}\right)^3 = -\frac{8}{27}.
x = -2. Divide by 5: x^3 = -8, so \displaystyle x = \sqrt[3]{-8} = -2.
x = -4. Subtract 20: x^3 = -64, so x = -4. Check: -64 + 20 = -44.
x = 4. The cube root of both sides gives x - 1 = 3, so x = 4. Check: (4 - 1)^3 = 27.
x^2 = 1 has two solutions, 1 and -1. x^3 = 1 has one, 1. x^2 = -1 has no real solution, since no square is negative. x^3 = -1 has one, -1.
8 along each edge and 64 on the top face. \displaystyle \sqrt[3]{512} = 8, and the top face is an 8 by 8 square of magnets.
16 cm. Counting: \displaystyle \sqrt[3]{1000} = 10 dice per edge, and 10 \times 1.6 = 16. Volume: each die is 1.6^3 = 4.096\ \text{cm}^3, the box is 1000 \times 4.096 = 4096\ \text{cm}^3, and \displaystyle \sqrt[3]{4096} = 16.
40 cm. 64 \times 1000 = 64\,000\ \text{cm}^3, and \displaystyle \sqrt[3]{64\,000} = 4 \times 10 = 40.
Between 7 and 8 cm, and more than 7.9 cm. 7^3 = 343 < 500 < 512 = 8^3, and 7.9^3 = 493.039 < 500, so the edge is between 7.9 and 8.
1350\ \text{cm}^2. The edge is \displaystyle \sqrt[3]{3375} = 15 cm, each face is 15^2 = 225\ \text{cm}^2, and six faces make 6 \times 225 = 1350.
\displaystyle \sqrt[4]{625} = 5 and \displaystyle \sqrt[5]{243} = 3, since 5^4 = 625 and 3^5 = 243. \displaystyle \sqrt[4]{-81} is not a real number, because a fourth power is a square and is never negative. \displaystyle \sqrt[5]{-1} = -1, since (-1)^5 = -1.
\displaystyle x^4 = \frac{1}{16} gives \displaystyle x = \frac{1}{2} or \displaystyle x = -\frac{1}{2}. x^5 = 32 gives x = 2 only. x^4 = -1 has no real solution.
Cubing keeps order: if a < b then a^3 < b^3, so two different numbers never share a cube and there is at most one solution. As x runs from negative to positive, x^3 passes through every value with no jumps (the graph argument above), so there is at least one. Squaring loses the sign, so a and -a share a square: a positive p gets two solutions, 0 gets one, and a negative p gets none, since no square is negative.
No. The right side is 2 + 3 = 5. The left side is \displaystyle \sqrt[3]{35}, and 3^3 = 27 < 35 < 64 = 4^3, so it lies between 3 and 4. A cube root of a sum is not the sum of the cube roots; the roots split only over products and quotients.
No. For x = -2, \displaystyle \sqrt[4]{(-2)^4} = \sqrt[4]{16} = 2, not -2. The even power wipes out the sign and the fourth root is never negative, so \displaystyle \sqrt[4]{x^4} = |x|. Odd powers keep the sign, which is why \displaystyle \sqrt[3]{x^3} = x for every x.