How do you factor $m^{3} + 64 n^{3}$ $m^3 + 64n^3$ ?

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Answer 1 · 2018-03-14T20:28:46

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

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:

$a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ $a^3 +b^3 = (a+b)(a^2-ab +b^2)$

Following the method shown we will have:

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

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:

How do you factor m3+64n3m^3 + 64n^3?