The cube root of an integer

Calculating with numbers: Calculating with powers and roots

The cube root of an integer

Every root that has been discussed so far was an example of a square root. There also exst higher roots>

We start with a special case, namely the cube root. Unlike a square root, you may take the cube root of a negative number.

The cube root

The cube root of a number $a\ge 0$ is by definition the number $w$ such that $w^3 = a$ . Notation: $w = \sqrt[3]{a}$ and $w=a^{\frac{1}{3}}$ .

Examples

$\begin{aligned}\sqrt[3]{8}&=\phantom{-}{2}\quad\text{because }2^3=8\text{ and }\\[0.2cm] \sqrt[3]{-8}&={-2}\quad\text{because }(-2)^3=-8\end{aligned}$

The calculation rules for cube roots resemble those of square roots.

Cube root of a cube

For any integer $n$ we have: $\sqrt[3]{n^3}=n$ and $\left(\sqrt[3]{n}\right)^3=n$

Example $n=4$

$\sqrt[3]{4^3}=4$ and $\left(\sqrt[3]{4}\right)^3=4$

Product rule of cube roots

For any integers $m$ and $n$ we have: $\sqrt[3]{m}\times \sqrt[3]{n}= \sqrt[3]{m\times n}$

example

$\sqrt[3]{2}\times \sqrt[3]{3}=\sqrt[3]{2\times 3}$ because $\begin{aligned}\left(\sqrt[3]{2}\times \sqrt[3]{3}\right)^3 &= \left(\sqrt[3]{2}\right)^3\times \left(\sqrt[3]{3}\right)^3\\ &= 2\times 3\\ &= 6\\ &= \left(\sqrt[3]{6}\right)^3 \\ &= \left(\sqrt[3]{2\times 3}\right)^3\end{aligned}$

How temping it may be, there is no sum rule for cube roots. For nonnegative numbers $m$ and $n$ we have: $\sqrt[3]{m}+\sqrt[3]{n}\;\red{\neq}\;\sqrt[3]{m+n}$ A concrete example with numbers illustrates this: $1=3-2=\sqrt[3]{27}+\sqrt[3]{-8}\;\red{\neq}\;\sqrt[3]{27-8}=\sqrt[3]{19}$

Quotient rule of cube roots

The cube root of a fraction with integers in the numerator and the denominator is equals to the quotient of the cube root of the numerator and the cube root of the denominator.

In formula language we have for integers $m$ and \(n): $\sqrt[3]{\frac{m}{n}}=\frac{\sqrt[3]{m}}{\sqrt[3]{n}}$

Examples

$\sqrt[3]{\frac{27}{64}}=\frac{\sqrt[3]{27}}{\sqrt[3]{64}}=\frac{3}{4}$ in indeed $\left(\frac{3}{4}\right)^3=\frac{3^3}{4^3}=\frac{27}{64}$

These rules can be used to simplify cube roots.

The cube root of a natural number greater than one, say $\sqrt[3]{n}$ , is called irreducible if $n$ has no cube number (i.e., a third power of a natural number) greater than 1 as divisor. Thus, $\sqrt[3]{35}=\sqrt[3]{5\times 7}$ and $\sqrt[3]{12}=\sqrt[3]{2^2\times 3}$ are irreducible cube roots, but $\sqrt[3]{40}$ is not irreducible, because $\begin{aligned}\sqrt[3]{40} &=\sqrt[3]{8\times 5}\\ &=\sqrt[3]{2^3\times 5}\\ &=\sqrt[3]{2^3}\times\sqrt[3]{5}\\ &=2\times \sqrt[3]{5}\end{aligned}$ The final expression we usually write shorter as $2\sqrt[3]{5}$ .

Each cube root of an integer not equal to 0 can be written in standard form, i.e., as an integer or as the product of an integer and an irreducible cube root.

The expression $m\sqrt[3]{n}$ for integers $m$ and $n>1$ is in standard form if there exist no cube number greater than 1 that divides $n$ .

You can find the standard form by 'extracting all third powers from the root'. The example below illustrate this.

Write the root $\;\sqrt[3]{-1125}\;$ in standard form.

First we find the(in absolute value) largest possible third power that divides $-1125$ and has the same sign as $-1125$ . In this case we can write: $\begin{aligned}-1125&=-125\times 9\\ \\ &={(-5)}^3\times 9\end{aligned}$ This follows, for example, from the prime factorisation of $1125$ : $1125=3^2\times 5^3$ Once the largest third power that divides $-1125$ is found we can apply the computational rules $\sqrt[3]{m\times n} = \sqrt[3]{m}\times\sqrt{n}$ and $\sqrt[3]{n^3}=n$ for integers $m$ and $n$ : $\begin{aligned} \sqrt[3]{-1125} &= \sqrt[3]{{(-5)}^3\times 9}\\ \\ &=\sqrt[3]{{(-5)}^3}\times \sqrt[3]{9}\\ \\ &=-5\sqrt[3]{9}\end{aligned}$

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