# Factoring a Sum of Cubes

## Problem

There is a box with a volume of $x^3-64$ cubic inches, a rectangular base, and a height of $x-4$ inches. What is the area of the box’s base? Use the formula $V=bh,$  where $V$ is volume, $B$ is area of the base, and $h$ is height. Give your answer as an expression in simplest form and in terms of $x.$

## Solution

You’re told that the box’s volume is equal to the area of its base times its height: $V=Bh.$ Then the area of this box’s base is equal to its volume divided by its height: $B=V/h.$

$B=\frac{V}{h}=\frac{x^3-64}{x-4}$

To get your answer into simplest form, factor the sum of cubes in the numerator and then carry out any possible cancellation between numerator and denominator. To do that factor the numerator. The numerator is the difference of two cubes, which factors like this:

$a^3-b^3=\left&space;(a-b&space;\right&space;)\left&space;(a^2+ab+b^2&space;\right&space;)$

Apply that to this case:

$a=x;&space;b=4$

Then

$x^3-4^3=\left&space;(&space;x-4\right&space;)\left&space;(&space;x^2+4x+16\right&space;)$

And the area of the base is

$\frac{\left&space;(x-4&space;\right&space;)\left&space;(&space;x^2+4x+4&space;\right&space;)}{x-4}=\bold{x^2+4x+16}$

The hard part of this problem is having the formula for the difference of cubes memorized. Alternatively, you can do polynomial long division or synthetic division. Synthetic division for this problem, $\frac{x^3-64}{x-4},$ should look like this:

Same result.

You may be asked about the sum, rather than the difference, of cubes. That formula is:

$a^3+b^3=\left&space;(&space;a+b\right&space;)\left&space;(&space;a^2-ab+b^2\right&space;)$

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