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

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Answer 1 · 2016-08-06T07:11:06

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

The sum of cubes identity can be written:

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

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

$54b^3+16$

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

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

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

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

How do you factor 54b^3+1654b^3+16?