How do you factor $2 a^{3} - 128$ $2a^3-128$ ?

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How do you factor $2 a^{3} - 128$ $2a^3-128$ ?

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Answer 1 · 2016-07-04T03:08:16

$(a - 4) (a^{2} + 4 a + 16)$ $(a - 4)(a^2 + 4a + 16)$

Explanation:

Apply the algebraic formula:
$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ $a^3 - b^3 = (a - b) (a^2 + ab + b^2)$
$y = 2 (a^{3} - 64) = 2 (a^{3} - 4^{3}) = 2 (a - 4) (a^{2} + 4 a + 16)$ $y = 2(a^3 - 64) = 2(a^3 - 4^3) = 2(a - 4)(a^2 + 4a + 16)$

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How do you factor $2 a^{3} - 128$ $2a^3-128$ ?

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