# How do you factor m^3 + 64n^3?

##### 1 Answer
Mar 14, 2018

${m}^{3} + 64 {n}^{3} = \left(m + 4 n\right) \left({m}^{2} - 4 m n + 16 {n}^{2}\right)$

#### Explanation:

This type of expression is known as the 'sum of two cubes', because both of the terms are perfect cubes.

Note the form of the factoring:

${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

Following the method shown we will have:

${m}^{3} + 64 {n}^{3} = \left(m + 4 n\right) \left({m}^{2} - 4 m n + 16 {n}^{2}\right)$

To factor the sum or difference of two cubes:

This is how it is done:

Make Two brackets,
The first is found from the cube root of each term.
The second bracket is formed from the first to get three terms:

• square the first term
• change the sign
• multiply the two terms together
• PLUS
• square the second term