Factorisation

**vuze88** · June 22nd 2010, 12:36 AM

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

**Dinkydoe** · June 22nd 2010, 01:30 AM

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

We can write it as a polynomial g with:

$c=x$

and

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

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

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

**Sudharaka** · June 22nd 2010, 01:52 AM

Originally Posted by vuze88

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

Dear vuze88,

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

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

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

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

Hope this helps.

**undefined** · June 22nd 2010, 02:09 AM

Originally Posted by Sudharaka

Dear vuze88,

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

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

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

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

Hope this helps.

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

Cheers.

**Sudharaka** · June 22nd 2010, 02:20 AM

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.

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