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)#