1. ## Factorisation

Factorize $\displaystyle a^4+b^4+c^2-2(a^2b^2+a^2c+b^2c)$

2. It seems like some obvious combination like $\displaystyle (\pm a^2\pm b^2\pm c)^2$ doesn't work for any combination of +/-

We can write it as a polynomial g with:

$\displaystyle c=x$

and

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

and then use abc-formula to find roots of $\displaystyle g(x)=0$
Then we can write $\displaystyle (x-r_1)(x-r_2)$ as a factorisation. (r_1,r_2 roots)

But such factorisation is not quite unique, in this case...

3. Originally Posted by vuze88
Factorize $\displaystyle a^4+b^4+c^2-2(a^2b^2+a^2c+b^2c)$
Dear vuze88,

Notice that, $\displaystyle (a^2+b^2-c)^2=a^4+b^4+c^2+2a^2b^2-2a^2c-2b^2c$

Therefore, $\displaystyle (a^2+b^2-c)^2-4a^2b^2=a^4+b^4+c^2-2a^2b^2-2a^2c-2b^2c$

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

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

Hope this helps.

4. Originally Posted by Sudharaka
Dear vuze88,

Notice that, $\displaystyle (a^2+b^2-c)^2=a^4+b^4+c^2+2a^2b^2-2a^2c-2b^2c$

Therefore, $\displaystyle (a^2+b^2-c)^2-4a^2b^2=a^4+b^4+c^2-2a^2b^2-2a^2c-2b^2c$

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

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

Hope this helps.
Really nice solution, just thought I should point out a small typo; the last "+" sign should be a "-".

Cheers.

5. Originally Posted by undefined
Really nice solution, just thought I should point out a small typo; the last "+" sign should be a "-".

Cheers.
Dear undefined,

Thanks. Corrected it.