# Thread: Sum of Two cubes

1. ## Sum of Two cubes

Hi I am having a few problems with a couple of questions of an algebra revision assignment. I have never heard of, nor been taught how to do the sum of two cubes up until this point.

a) 54x^3 +2
b) x^6 -7x^3 -8

I have been having a go with the first one as it seems simpler. Using the theory I have found on wikipedia . a^3 + b^3 = (a+b)(a^2 -2ab + b^2) I have tried to *fudge* an answer but it isnt working out when I expand it to check my work.

a) 2(27x^3 +1)
2(1+3x)(9x^2 -6y+1)

This is where I come to trying to jam it into a formula, however through not knowing what to do or making a blind mistake it doesnt seem to be correct.
I have no idea where to start with the second problem, seeing as I cannot do a much easier question. Any help is much appreciated

2. Hello, Webby!

You are expected to know how to factor the sum or difference of two cubes.

. . . . . $\displaystyle a^3 + b^3 \;=\;(a + b)(a^2 - ab + b^2)$ .(1)

. . . . . $\displaystyle a^3-b^3 \;=\;(a-b)(a^2 + ab +b^2)$ .(2)

Factor: .$\displaystyle a)\;\; 54x^3 +2$

$\displaystyle \text{First, factor out 2: }\;2\underbrace{\left(27x^3 + 1\right)}_{\text{sum of cubes}}$

$\displaystyle \text{We have: }\;27x^3 + 1 \;=\;(3x)^3 + (1)^3$

. . $\displaystyle \text{So: }\:a = 3x,\;b = 1$

. . $\displaystyle \text{Substitute into {\color{blue}(1)}: }\;(3x+1)(9x^2 - 3x + 1)$

$\displaystyle \text{Answer: }\;2(3x+1)(9x^2-3x + 1)$

$\displaystyle b)\;\;x^6 -7x^3 -8$

$\displaystyle \text{Factor: }\;(x^3 + 1)(x^3-8)$

$\displaystyle \text{We have: }\;\begin{Bmatrix}x^3+1 &=& (x)^3 + (1)^3 && \text{sum of cubes} \\ x^3-8 &=& (x)^3 - (2)^3 && \text{diff. of cubes} \end{Bmatrix}$

$\displaystyle \text{Hence: }\;\begin{array}{ccccc}{\color{blue}(1)} & x^3+1^3 &=& (x+1)(x^2-x+1) \\ {\color{blue}(2)} & x^3 - 2^3 &=& (x-2)(x^2 + 2x + 1) \end{array}$

$\displaystyle \text{Answer: }\;(x+1)(x^2-x+1)(x-2)(x^2+2x+1)$