How do you factor $54 b^{3} + 16$ $54b^3+16$ ?

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How do you factor $54 b^{3} + 16$ $54b^3+16$ ?

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Answer 1 · 2016-08-06T07:11:06

$54 b^{3} + 16 = 2 (3 b + 2) (9 b^{2} - 6 b + 4)$ $54b^3+16=2(3b+2)(9b^2-6b+4)$

Explanation:

The sum of cubes identity can be written:

$x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2})$ $x^3+y^3 = (x+y)(x^2-xy+y^2)$

Note that both $54$ $54$ and $16$ $16$ are divisible by $2$ $2$ , so we find:

$54 b^{3} + 16$ $54b^3+16$

$= 2 (27 b^{3} + 8)$ $=2(27b^3+8)$

$= 2 ({(3 b)}^{3} + 2^{3})$ $=2((3b)^3+2^3)$

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

$= 2 (3 b + 2) (9 b^{2} - 6 b + 4)$ $=2(3b+2)(9b^2-6b+4)$

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How do you factor $54 b^{3} + 16$ $54b^3+16$ ?

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How do you factor $54 b^{3} + 16$ $54b^3+16$ ?