how do you solve 64+27u^6

Question

Factoring sums and differences of cubes

Answer 1

This can be written 4^3+(3u^2)^3. X^3+Y^3=(X+Y)(X^2-XY+Y^2), so putting X=4 and Y=3u^2 we get:

(4+3u^2)(16-12u^2+9u^4)=64-48u^2+36u^4+48u^2-36u^4+27u^6=64+27u^6.

Answer 2

Given 64+27u^6
factors of the equation
(3u^2+4)(3u^2-6u+4)(3u^2+6u+4)
u=±2i/√3 or u=1/3(±3-i√3) or u=1/3(-3+i√3)

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