How do you factor $27 a^{3} - 64 b^{3}$ $27a^3-64b^3$ ?

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How do you factor $27 a^{3} - 64 b^{3}$ $27a^3-64b^3$ ?

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Answer 1 · 2016-01-08T12:10:10

$27 a^{3} - 64 b^{3} = (3 a - 4 b) (9 a^{2} + 12 a b + 16 b^{2})$ $27a^3-64b^3=(3a-4b)(9a^2+12ab+16b^2)$

Explanation:

Remembering that:

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

we can try to write

$27 a^{3} - 64 b^{3}$ $27a^3-64b^3$

like a difference of cubes

$27 a^{3} - 64 b^{3} = 3^{3} a^{3} - 2^{6} b^{3} = 3^{3} a^{3} - {(2^{2})}^{3} b^{3} =$ $27a^3-64b^3=3^3a^3-2^6b^3=3^3a^3-(2^2)^3b^3=$
${(3 a)}^{3} - 4^{3} b^{3} = {(3 a)}^{3} - {(4 b)}^{3}$ $(3a)^3-4^3b^3=(3a)^3-(4b)^3$

Now we can apply the rule:

$27 a^{3} - 64 b^{3} = {(3 a)}^{3} - {(4 b)}^{3} =$ $27a^3-64b^3=(3a)^3-(4b)^3=$
$= (3 a - 4 b) ({(3 a)}^{2} + 12 a b + {(4 b)}^{2})$ $=(3a-4b)((3a)^2+12ab+(4b)^2)$
$= (3 a - 4 b) (9 a^{2} + 12 a b + 16 b^{2})$ $=(3a-4b)(9a^2+12ab+16b^2)$

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How do you factor $27 a^{3} - 64 b^{3}$ $27a^3-64b^3$ ?

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How do you factor $27 a^{3} - 64 b^{3}$ $27a^3-64b^3$ ?