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

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Answer 1 · 2018-03-14T20:28:46

#m^3 +64n^3 = (m+4n)(m^2-4mn +16n^2)#

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 = (a+b)(a^2-ab +b^2)#

Following the method shown we will have:

#m^3 +64n^3 = (m+4n)(m^2-4mn +16n^2)#

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: