How do you factor $27k^3+64d^3$ ?

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Answer 1 · 2016-08-29T08:50:56

$27k^3+64d^3=(3k+4d)(9k^2-12kd+16d^2)$

The sum of cubes identity can be written:

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

We can use this with $a=3k$ and $b=4d$ to find:

$27k^3+64d^3$

$=(3k)^3+(4d)^3$

$=(3k+4d)((3k)^2-(3k)(4d)+(4d)^2)$

$=(3k+4d)(9k^2-12kd+16d^2)$

That is as far as we can go with Real coefficients. If we allow Complex coefficients then this can be factored further as:

$=(3k+4d)(3k+4omegad)(3k+4omega^2d)$

where $omega = -1/2+sqrt(3)/2i$ is the primitive Complex cube root of $1$

Answer 2 · 2016-08-29T09:07:02

= $(3k +4d)(9k^2 -12dk +16d^2)$

This difficulty is recognising that the values are all cubes.

This expression is the sum of cubes.

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

$27k^3 +64d^3 = (3k +4d)(9k^2 -12dk +16d^2)$

How do you factor 27k^3+64d^327k^3+64d^3?