How do you factor $b^{3} - 8$ $b^3 - 8$ ?

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How do you factor $b^{3} - 8$ $b^3 - 8$ ?

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Answer 1 · 2018-03-07T12:25:15

$(b - 2) (b^{2} + 2 b + 4)$ $(b-2)(b^2+2b+4)$

Explanation:

$b^{3} - 8$ $b^3-8$
$\Rightarrow {(b)}^{3} - {(2)}^{3}$ $rArr (b)^3-(2)^3$
$\Rightarrow {(b - 2)}^{3} + 3 \cdot b \cdot 2 (b - 2)$ $rArr (b-2)^3+3*b*2(b-2)$
$\Rightarrow (b - 2) [{(b - 2)}^{2} + 6 b]$ $rArr (b-2)[(b-2)^2+6b]$
$\Rightarrow (b - 2) [b^{2} - 4 b + 4 + 6 b]$ $rArr (b-2)[b^2-4b+4+6b]$
$\Rightarrow (b - 2) (b^{2} + 2 b + 4)$ $rArr (b-2)(b^2+2b+4)$

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How do you factor $b^{3} - 8$ $b^3 - 8$ ?

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