How do you factor $192 u^{3} - 81$ $192u^3 - 81$ ?

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How do you factor $192 u^{3} - 81$ $192u^3 - 81$ ?

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Answer 1 · 2018-03-23T15:09:58

$3 (16 u^{2} + 12 u + 9)$ $3(16u^2+12u+9)$

Explanation:

$3 (64 u^{3} - 27)$ $3(64u^3-27)$
$3 [{(4 u)}^{3} - {(3)}^{3})]$ $3[(4u)^3-(3)^3)]$
We know that $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ $a^3-b^3=(a-b)(a^2+ab+b^2)$
In this case, $a = 4 u$ $a=4u$ and $b = 3$ $b=3$
Therefore:
${(4 u)}^{3} - {(3)}^{3} = (4 u - 3) (16 u^{2} + 12 u + 9)$ $(4u)^3-(3)^3=(4u-3)(16u^2+12u+9)$
$3 (16 u^{2} + 12 u + 9)$ $3(16u^2+12u+9)$ (we still have the 3 leftover from before)

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How do you factor $192 u^{3} - 81$ $192u^3 - 81$ ?

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