Cube Roots and Cube Root Equations | Grade 8 Math | LevelKite

You already know how to work backward from a perfect square. In the last lesson, you read $\sqrt{49} = 7$ as the answer to a question: which nonnegative number (zero or positive) has a square of $49$ ? In this lesson, you will ask the same kind of question about cubes.

To cube a number means to use it as a factor three times. For example, $4^3 = 4 \cdot 4 \cdot 4 = 16 \cdot 4 = 64$ . The name comes from geometry: a cube whose edges are each $4$ centimeters long has a volume of $4 \cdot 4 \cdot 4 = 64$ cubic centimeters.

Now run that backward. Suppose a cube has a volume of $64$ cubic centimeters. How long is each edge? You need the number whose cube is $64$ . That number is $4$ , so each edge is $4$ centimeters long. You call $4$ the cube root of $64$ .

The cube root of a real number $n$ is the real number whose cube is $n$ . (A real number is any number on the number line: positive, negative, or zero.) You write the cube root of $n$ as $\sqrt[3]{n}$ . The small $3$ on the symbol tells you it is a cube root, not a square root.

So $\sqrt[3]{64} = 4$ because $4^3 = 64$ , and $\sqrt[3]{8} = 2$ because $2^3 = 2 \cdot 2 \cdot 2 = 8$ .

A perfect cube is the cube of an integer. The numbers $8$ and $64$ are perfect cubes. So is $-8$ , because $(-2)(-2)(-2) = -8$ . To find the cube root of a perfect cube, ask yourself: which integer was cubed? The table lists the perfect cubes you will meet most often.

Cubes of the integers from 1 to 10 and of their opposites

Integer $a$	Cube $a^3$	Cube root fact	Opposite integer $-a$
$1$	$1 \cdot 1 \cdot 1 = 1$	$\sqrt[3]{1} = 1$	$(-1)^3 = -1$ , so $\sqrt[3]{-1} = -1$
$2$	$2 \cdot 2 \cdot 2 = 8$	$\sqrt[3]{8} = 2$	$(-2)^3 = -8$ , so $\sqrt[3]{-8} = -2$
$3$	$3 \cdot 3 \cdot 3 = 27$	$\sqrt[3]{27} = 3$	$(-3)^3 = -27$ , so $\sqrt[3]{-27} = -3$
$4$	$4 \cdot 4 \cdot 4 = 64$	$\sqrt[3]{64} = 4$	$(-4)^3 = -64$ , so $\sqrt[3]{-64} = -4$
$5$	$5 \cdot 5 \cdot 5 = 125$	$\sqrt[3]{125} = 5$	$(-5)^3 = -125$ , so $\sqrt[3]{-125} = -5$
$6$	$6 \cdot 6 \cdot 6 = 216$	$\sqrt[3]{216} = 6$	$(-6)^3 = -216$ , so $\sqrt[3]{-216} = -6$
$7$	$7 \cdot 7 \cdot 7 = 343$	$\sqrt[3]{343} = 7$	$(-7)^3 = -343$ , so $\sqrt[3]{-343} = -7$
$8$	$8 \cdot 8 \cdot 8 = 512$	$\sqrt[3]{512} = 8$	$(-8)^3 = -512$ , so $\sqrt[3]{-512} = -8$
$9$	$9 \cdot 9 \cdot 9 = 729$	$\sqrt[3]{729} = 9$	$(-9)^3 = -729$ , so $\sqrt[3]{-729} = -9$
$10$	$10 \cdot 10 \cdot 10 = 1000$	$\sqrt[3]{1000} = 10$	$(-10)^3 = -1000$ , so $\sqrt[3]{-1000} = -10$

Two patterns in the table explain how cube roots behave.

A cube keeps the sign. Cubing a positive number gives a positive result, because every factor is positive. Cubing a negative number gives a negative result: in $(-2)(-2)(-2)$ , the first two factors multiply to $4$ , and then $4 \cdot (-2) = -8$ . Also, $0^3 = 0$ . So the cube of a real number always has the same sign as the number itself.

Different numbers have different cubes. A larger real number always has a larger cube. For example, $2 < 3$ and $8 < 27$ . In the same way, $-3 < -2$ and $-27 < -8$ . So two different real numbers never share the same cube.

These two patterns lead to the main idea of this lesson.

Key concept

Every real number has exactly one real cube root

For every real number $n$ , exactly one real number has a cube equal to $n$ . That number is the cube root of $n$ , written $\sqrt[3]{n}$ .

The cube root has the same sign as $n$ : it is positive when $n > 0$ , negative when $n < 0$ , and $0$ when $n = 0$ . For example, $\sqrt[3]{27} = 3$ and $\sqrt[3]{-27} = -3$ .

In the formula, $n$ and $r$ stand for real numbers.

\sqrt[3]{n} = r \quad \text{exactly when} \quad r^3 = n

Worked example

Find the cube root of a negative perfect cube

Find $\sqrt[3]{-216}$ by reasoning, without looking it up in the table.

Turn the symbol into a question: which real number has a cube of negative 216? By the definition of a cube root, the answer cubed must equal negative 216.
$\left(\sqrt[3]{-216}\right)^3 = -216$
Decide the sign. The cube of a real number has the same sign as the number itself. Negative 216 is a negative number, so its cube root is negative.
Find the size by looking for the positive integer whose cube is 216. Test 5: you get 125, which is too small. A larger number has a larger cube, so the integer you need is greater than 5.
$5^3 = 5 \cdot 5 \cdot 5 = 25 \cdot 5 = 125$
Test 6. You get exactly 216, so the cube root you want should be negative 6.
$6^3 = 6 \cdot 6 \cdot 6 = 36 \cdot 6 = 216$
Check by cubing negative 6. The first two factors multiply to positive 36, and 36 times negative 6 is negative 216. Each real number has only one real cube root, so negative 6 is the answer.
$(-6)^3 = (-6)(-6)(-6) = 36 \cdot (-6) = -216$

Answer

The cube root of $-216$ is $-6$ . In symbols, $\sqrt[3]{-216} = -6$ because $(-6)^3 = -216$ .

Now use cube roots to solve equations. In an equation such as $x^3 = 64$ , the letter $x$ stands for a real number that you do not know yet. A solution is a value of $x$ that makes the equation true.

Read the equation as a question: which real number has a cube of $64$ ? That is exactly the question a cube root answers. Every real number $n$ has exactly one real cube root, so the equation $x^3 = n$ has exactly one real solution, $x = \sqrt[3]{n}$ . Like every cube root, that solution has the same sign as $n$ , and it is $0$ when $n = 0$ .

Worked example

Solve an equation with a perfect cube

Solve $x^3 = 64$ , where $x$ stands for a real number.

Take the cube root of both sides. Each real number has exactly one real cube root, so equal numbers have equal cube roots. That is why this step keeps the equation true.
$\sqrt[3]{x^3} = \sqrt[3]{64}$
Simplify the left side. The cube root of x cubed is x, because x is the one real number whose cube is x cubed. This holds whether x is positive, negative, or zero.
$x = \sqrt[3]{64}$
Evaluate the cube root. Because 64 is positive, its cube root is positive. The table shows that 64 is the cube of 4.
$x = 4$
Check by substituting 4 for x in the original equation. The left side equals 64, so 4 is a solution. A different real number would have a different cube, so 4 is the only real solution.
$4^3 = 4 \cdot 4 \cdot 4 = 16 \cdot 4 = 64$

Answer

The equation $x^3 = 64$ has exactly one real solution, $x = 4$ .

Don't borrow the plus-or-minus sign from squares. In the last example, a common mistake is to write $x = \pm 4$ , because $x^2 = 64$ has two solutions, $8$ and $-8$ . Test $-4$ in $x^3 = 64$ : $(-4)^3 = (-4)(-4)(-4) = 16 \cdot (-4) = -64$ , not $64$ . The mistake fails because squaring removes a negative sign, as in $(-8)(-8) = 64$ , but cubing keeps it.

Getting two solutions is a pattern for square equations, and only for positive numbers: for real numbers $x$ and $n$ , the equation $x^2 = n$ has two solutions, $\sqrt{n}$ and $-\sqrt{n}$ , when $n > 0$ . It has one solution, $0$ , when $n = 0$ , and no real solution when $n < 0$ . A cube equation $x^3 = n$ has exactly one real solution for every real number $n$ , so $x^3 = 64$ gives only $x = 4$ .

Worked example

Solve an equation with a negative fraction

Solve $x^3 = -\frac{8}{125}$ , where $x$ stands for a real number.

Take the cube root of both sides, as in the last example. The equation has exactly one real solution. That solution is negative, because the right side is negative.
$x = \sqrt[3]{-\frac{8}{125}}$
Find the size. Cubing a fraction multiplies three copies of it, so the three numerators multiply together and the three denominators multiply together. You need a numerator whose cube is 8 and a denominator whose cube is 125. They are 2 and 5, so the solution should be negative 2/5.
$2 \cdot 2 \cdot 2 = 8 \qquad 5 \cdot 5 \cdot 5 = 125$
Check by cubing negative 2/5. The first two factors multiply to positive 4/25, and 4/25 times negative 2/5 is negative 8/125. That matches the right side of the original equation.
$\left(-\frac{2}{5}\right)\left(-\frac{2}{5}\right)\left(-\frac{2}{5}\right) = \frac{4}{25} \cdot \left(-\frac{2}{5}\right) = -\frac{8}{125}$

Answer

The equation $x^3 = -\frac{8}{125}$ has exactly one real solution, $x = -\frac{2}{5}$ .

Takeaway. For every real number $n$ , the cube root $\sqrt[3]{n}$ is the one real number whose cube is $n$ , and it has the same sign as $n$ (it is $0$ when $n = 0$ ). That is why the equation $x^3 = n$ , where $x$ stands for a real number, has exactly one real solution, $x = \sqrt[3]{n}$ , and never a plus-or-minus pair. When $n$ is a perfect cube, you can find that solution by asking which integer was cubed.

Next, you will learn that $\sqrt{2}$ is irrational, which means that no fraction $\frac{p}{q}$ with integers $p$ and $q$ and $q \neq 0$ is equal to it.