Factorising

GerCoffey · **Joined:** Tue, 14 Sep 2010 09:02:34 UTC **Posts:** 3

Can Somebody explain to me how you would solve the following question, thank you.

Factorise fully the following expression:
x3 +y3 + 3x + 3y

outermeasure · **Posted:** Tue, 14 Sep 2010 09:39:36 UTC

GerCoffey wrote:

Can Somebody explain to me how you would solve the following question, thank you.

Factorise fully the following expression:
x3 +y3 + 3x + 3y

, etc. so

.

Or do you mean x^3= x^3

, etc.? If so, you cannot factorise x^3+y^3+3^x+3^y

in any meaningful way.

GerCoffey · **Joined:** Tue, 14 Sep 2010 09:02:34 UTC **Posts:** 3

Thanks for the response, sorry for not being clear, the first two are cubes so it reads

x^3 + y^3 + 3x + 3y

Soroban · **Posted:** Tue, 14 Sep 2010 13:52:52 UTC

Hello, GerCoffey!

Quote:

Factor: .

Note that the first two terms is a "sum of cubes".
And you should know that: . $a^3 + b^3 \:=\:(a+b)(a^2-ab+b^2)$

So we have: . (x+y)(x^2-xy + y^2) + 3(x+y)

$\text{We note that we have: }\;\underbrace{(x+y)}_{\text{common factor}}\!\!\!\!(x^2-xy+y^2) + 3\!\!\!\!\!\!\underbrace{(x+y)}_{\text{common factor}}$

Factor out the common $(x+y)\!:\quad (x+y)\,(x^2-xy + y^2 + 3)$

GerCoffey · **Joined:** Tue, 14 Sep 2010 09:02:34 UTC **Posts:** 3

Thank you, one of those ones that now I know is so easy but couldnt for the life of me see it. such a relief.

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Factorising

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