Cube Roots and Higher Order Roots

Cube Roots
and
Other Radicals

Square roots ask what squared is the radicand. A geometric explanation is that given the area of a square, what’s the side length? A geometric explanation of a cube root is given the volume of a cube, what’s the side length. The way you find the volume of a cube is multiply the length by itself three times (cube it).

The way we write cube root is similar to square roots, with one very big difference, the index.

$\begin{array}{l}\sqrt{a}\to squareroot\\ \sqrt[3]{a}\to cuberoot\end{array}$

There actually is an index for a square root, but we don’t write the two. It is just assumed to be there.

Warning: When writing cube roots, or other roots, be careful to write the index in the proper place. If not, what you will write will look like multiplication and you can confuse yourself. When writing by hand, this is an easy thing to do.

$3\sqrt{8}\sqrt[3]{8}$

To simplify a square root you factor the radicand and look for the largest perfect square. To simplify a cubed root you factor the radicand and find the largest perfect cube. A perfect cube is a number times itself three times. The first ten are 1, 8, 27, 64, 125, 216, 343, 512, 729, 1,000.

Let’s see an example:

Simplify:

$\sqrt[3]{16}$

Factor the radicand, 16, find the largest perfect cube, which is 8.

$\sqrt[3]{8}\times \sqrt[3]{2}$

The cube root of eight is just two.

$2\sqrt[3]{2}$

The following is true,

$\sqrt[3]{16}=2\sqrt[3]{2}$,

only if

${\left(2\sqrt[3]{2}\right)}^3=16$

Arithmetic with other radicals, like cube roots, work the same as they do with square roots. We will multiply the rational numbers together, then the irrational numbers together, and then see if simplification can occur.

${\left(2\sqrt[3]{2}\right)}^3=2^3\times {\left(\sqrt[3]{2}\right)}^3$

Two cubed is just eight and the cube root of two cubed is the cube root of eight.

$2^3\times {\left(\sqrt[3]{2}\right)}^3=8\times \left(\sqrt[3]{2}\right)\left(\sqrt[3]{2}\right)\left(\sqrt[3]{2}\right)$

$2^3\times {\left(\sqrt[3]{2}\right)}^3=8\times \sqrt[3]{2\cdot 2\cdot 2}$

$2^3\times {\left(\sqrt[3]{2}\right)}^3=8\times \sqrt[3]{8}$

The cube root of eight is just two.

$8\times \sqrt[3]{8}=8\times 2$

${\left(2\sqrt[3]{2}\right)}^3=16$

Negatives and cube roots: The square root of a negative number is imagery. There isn’t a real number times itself that is negative because, well a negative squared is positive. Cubed numbers, though, can be negative.

$-3\times -3\times -3=-27$

So the cube root of a negative number is, well, a negative number.

$\sqrt[3]{-27}=-3$

Other indices (plural of index): The index tells you what power of a base to look for. For example, the 6^th root is looking for a perfect 6^th number, like 64. Sixty four is two to the sixth power.

$\sqrt[6]{64}=2\text{}\text{}because{2}^6=64.$

A few points to make clear.

·         If the index is even and the radicand is negative, the number is irrational.

·         If the radicand does not contain a factor that is a perfect power of the index, the number is irrational

·         All operations, including rationalizing the denominator, work just as they do with square roots.

Rationalizing the Denominator:

Consider the following:

$\frac{9}{\sqrt[3]{3}}$

If we multiply by the cube root of three, we get this:

$\frac{9}{\sqrt[3]{3}}\cdot \frac{\sqrt[3]{3}}{\sqrt[3]{3}}=\frac{9\sqrt[3]{3}}{\sqrt[3]{9}}$

Since 9 is not a perfect cube, the denominator is still irrational. Instead, we need to multiply by the cube root of nine.

$\frac{9}{\sqrt[3]{3}}\cdot \frac{\sqrt[3]{9}}{\sqrt[3]{9}}=\frac{9\sqrt[3]{9}}{\sqrt[3]{27}}$

Since twenty seven is a perfect cube, this can be simplified.

$\frac{9}{\sqrt[3]{3}}=\frac{9\sqrt[3]{9}}{3}$

And always make sure to reduce if possible.

$\frac{9}{\sqrt[3]{3}}=3\sqrt[3]{9}$

This is a bit tricky, to be sure. The way the math is written does not offer us a clear insight into how to manage the situation. However, the topic we will see next, rational exponents, will make this much clearer.

 

Practice Problems:

Simplify or perform the indicated operations:

$\begin{array}{l}1.\sqrt[4]{64}\\ \\ \\ 2.\sqrt[3]{9}+4\sqrt[3]{9}\\ \\ \\ \\ 3.\sqrt[3]{9}\times 4\sqrt[3]{9}\\ \\ \\ 4.\sqrt[5]{64}\\ \\ \\ 5.\frac{\sqrt{7}}{\sqrt[3]{7}}\end{array}$