1. ## Factorisation

Hi How would I factorise this:

a^6 + b^6

I change it into a cubic so,

(a^3)^2 + (b^3)^2 then ?

And how would I fully factorise this: x^3 - 3x^2 - 4

Thanks!

2. Originally Posted by classicstrings
Hi How would I factorise this:
a^6 + b^6
...
Hello, classicstring,

I can't offer you a more systematic way to do your problem. I did a little bit trial and error and long division and found this product:

$\displaystyle a^6+b^6=(a^2+b^2)(a^4-a^2b^2+b^4)$

Bye

EB

3. Originally Posted by earboth
Hello, classicstring,

I can't offer you a more systematic way to do your problem. I did a little bit trial and error and long division and found this product:

$\displaystyle a^6+b^6=(a^2+b^2)(a^4-a^2b^2+b^4)$

Bye

EB
Another way to do it will give you...
$\displaystyle a^6+b^6=-1\left(\left(a^3-b^3\right)\left(b^3-a^3\right)\right)$

4. I just realised that I think it is like cubic expansion, just doubling the powers.

Any ideas on this one?

x^3 - 3x^2 - 4

I think there are two nonreal solns and one real one.

Thanks!

5. Originally Posted by classicstrings
I just realised that I think it is like cubic expansion, just doubling the powers.
Elementry, my Dear classicstrings.
Use,
$\displaystyle x^3+y^3=(x+y)(x^2-xy+y^2)$
Thus,
$\displaystyle a^6+b^6=(a^2)^3+(b^2)^3$

6. Originally Posted by ThePerfectHacker
Elementry, my Dear classicstrings.
Use,
$\displaystyle x^3+y^3=(x+y)(x^2-xy+y^2)$
Thus,
$\displaystyle a^6+b^6=(a^2)^3+(b^2)^3$
I like your panache PerfectHacker, the explanation totally clears things up, - EDIT: yeah I was confused a bit as I had wrote

$\displaystyle a^6+b^6=(a^3)^2+(b^3)^2$

Which is wrong!

By any chance can you help me with factorising this one?

x^3 - 3x^2 - 4