How do you factor $2 w^{3} + 54$ $2w^3 + 54$ ?

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Answer 1 · 2015-12-19T01:49:38

$2 (w + 3) (w^{2} - 3 w + 9)$ $2(w+3)(w^2-3w+9)$

First, factor out a common $2$ $2$ .

$= 2 (w^{3} + 27)$ $=2(w^3+27)$

Notice that $(w^{3} + 27)$ $(w^3+27)$ is a sum of cubes, which follows the rule

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

Thus,

$2 (w^{3} + 27) = 2 ({(w)}^{3} + {(3)}^{3}) = 2 (w + 3) (w^{2} - 3 w + 9)$ $2(w^3+27)=2((w)^3+(3)^3)=2(w+3)(w^2-3w+9)$

How do you factor 2w^3 + 542w3+542w^3 + 54?