Tricky Decomposition Decompose the following polynomial:
$A=x^4(y^2+z^2) + y^4(z^2+x^2) + z^4(x^2+y^2) +2x^2 y^2 z^2 $
(by the way I'm not quite sure about the tags and the title feel free to edit them if you wish)
 A: $$
A = x^4(y^2+z^2) + x^2(y^4+2y^2 z^2 +z^4) +y^2z^2(y^2+z^2) \\
= (y^2+z^2)\left(x^4+(y^2+z^2)x^2 +y^2z^2\right) \\
=(y^2+z^2)(x^2+y^2)(z^2+x^2)
$$ 
A: Firstly, let $x^2=a$, let $y^2=b$ and let $z^2=c$ then we have $a^2b+a^2c+b^2a+b^2c+c^2a+c^2b+2abc$. I'm afraid that I just recognise this, just like how I recognise the difference of two squares, as:
$$(a+b)(b+c)(a+c)$$
A: Define the polynomial:
$$
f(a,b,c) = a^2(b + c) + b^2(c + a) + c^2(a + b) + 2abc
$$
Notice that $f$ is cyclic:
$$
f(a,b,c) = f(b,c,a) = f(c,a,b)
$$
Observe that since:
\begin{align*}
f(a,-a,c)
&= a^2(-a + c) + a^2(c + a) + 0 - 2a^2c \\
&= a^2(-a + c + c + a - 2c) \\
&= 0
\end{align*}
it follows by the Factor Theorem that $a+b$ is a factor of $f$. But since $f$ is cyclic, so are $b+c$ and $a+c$. Hence, for some constant $k$, we have:
$$
a^2(b + c) + b^2(c + a) + c^2(a + b) + 2abc = k(a + b)(b + c)(a + c)
$$
By comparing the coefficient of $a^2b$, observe that $k = 1$. Hence, we conclude that:
$$
A = f(x^2, y^2, z^2) = (x^2 + y^2)(y^2 + z^2)(x^2 + y^2)
$$
