2009/030) Factorize the polynomial 2{(bc)^2+(ca)^2+(ab)^2} - {a^4+b^4+c^4}

Factorize the polynomial
2{(bc)^2+(ca)^2+(ab)^2} - {a^4+b^4+c^4}?

We need to factorize
{a^4+b^4+c^4} – 2(bc)^2 - 2(ca)^2 - 2(ab)^2 and multiply by -1
Now
{a^4+b^4+c^4} – 2(bc)^2 - 2(ca)^2 - 2(ab)^2
= a^4 – 2a^2(b^2+c^2) + b^4+c^4-2ab^2c^2
= a^4 – 2a^2(b^2+c^2) + b^4+c^4+2b^2c^2- 4b^2c^2
= a^4 – 2a^2(b^2+c^2) + (b^2+c^2)^2- 4b^2c^2
= (a^2-b^2-c^2)^2 – (2bc)^2
= (b^2+c^2 – a^2)^2- (2bc)^2
= (b^2+c^2-a^2+2bc)(b^2+c^2-a^2-2bc)
= (b^2+c^2+2bc-a^2)(b^2+c^2-2bc-a^2)
= ((b+c)^2-a^2)((b-c)^2 – a^2)
= (b+c+a)(b+c-a)(b-c+a)(b-c-a)

So given expression
2{(bc)^2+(ca)^2+(ab)^2} - {a^4+b^4+c^4} = - (b+c+a)(b+c-a)(b-c+a)(b-c-a)
= (a+b+c))b+c-a)(a+b-c)(a+c-b) multiplying last term by -1 to get symmetrical form

2009/029) prove a+b+c=0 ==> 2a⁴ +2b⁴ +2c⁴ = n²

a+b+c = 0

the a+b = - c
square both sides

(a^2+b^2+2ab) = c^2
so (a^2+b^2) = c^2 -2ab

now again square both sides

a^4+b^4 + 2a^2b^2 = (c^2-2ab)^2
or a^4 + b^4 + 2a^2b^2 = c^4 + 4a^2b^2 - 4abc^2
or a^4+b^4 = c^4 + 2a^2b^2 - 4abc^2
add c^4 on both sides to get

a^4+b^4+c^4 = 2c^4 + 2a^2b^2 - 4abc^2

or 2(a^4+b^4+c^4) = 4c^4+4a^2b^2 - 8abc^2
= 4(c^4-2abc^2+a^2b^2)
= 4(c^2-ab)^2

which is a perfect quare putting n = 2 | c^2-ab|
proved