how to factorize $(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)$?

Question

how to factorize $(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)$?

this is one of my hard questions.

I know it is related to $(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc$

but I don't know how to factorize it.

Answer 1

The solution is:

$$\begin{align}(a^2 + b^2 + c^2)^2 - 2(a^4 + b^4 + c^4) & = -a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2-c^4 \\ & = -(a^4 - 2a^2b^2 - 2a^2c^2 + b^4-2b^2c^2+c^4)\\ & = -(a^2-2ac-b^2+c^2)(a^2 + 2ac-b^2+c^2)\\ & = -(a^2-ac - ac - b^2 + c^2 + (ab - ab))(a^2 + 2ac-b^2+c^2)\\ & = -(a^2-ab-ac+ab-b^2-bc-ac+bc+c^2)(a^2 + 2ac-b^2+c^2)\\ & = -(a(a-b-c) + b(a-b-c) - c(a-b-c))(a^2 + 2ac-b^2+c^2)\\& = -(a-b-c)[a+b-c](a^2 + 2ac-b^2+c^2) \\ & =-(a-b-c)(a+b+c)(a^2-ab+ac+ab-b^2 + bc + ac - bc + c^2)\\ & = -(a-b-c)(a+b+c)(a(a-b+c)+b(a-b+c)+c(a-b+c))\\& = -(a-b-c)(a+b+c)(a-b+c)[a+b+c]\\ & =-(a-b-c) (a+b-c) (a-b+c) (a+b+c)\end{align}$$

The key to this factorization was to add and subtract a quantity $ab$ in both instances and group things together.

Answer 2

In a formal software it's: $$(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)=(a+b-c)(a-b+c)(-a+b+c)(a+b+c) $$

Answer 3

You can use the factor theorem. We can treat the expression as a function of $a$. You can replace $a$ with $x$ if you're more comfortable using that symbol for the variable of a function. I won't do that. Then we can ask Wolfram Alpha to "solve (a^2+b^2+c^2)^2−2(a^4+b^4+c^4) = 0 for a". By the way, you can do this by hand using the quadratic formula because the expression is a quadratic in $a^2$.

It says that the roots (solutions in $a$) are: $a = \pm(b + c)$ and $a = \pm(c-b)$. So by the factor theorem we get $(a+b−c)(a−b+c)(−a+b+c)(a+b+c)$.