Not so simple as sub and sub

First simplify the denominator:

$a^2b^2+c^2(a^2+b^2)=a^2b^2+a^2c^2+b^2c^2=(ab+ac+bc)^2-2abc(a+b+c)$ But $a+b+c=0$ .So $2abc(a+b+c)=2abc(0)=0$ and we are left with $(ab+ac+bc)^2$ Now the numerator:

$a^4+b^4+c^4$ can be written as $(a^2+b^2+c^2)^2-2(a^2b^2+a^2c^2+b^2c^2)$ .

Which can be written as $((a+b+c)^2-2(ab+ac+bc))^2-2(a^2b^2+a^2c^2+b^2c^2)$

But we know that $a^2b^2+a^2c^2+b^2c^2=(ab+ac+bc)^2$ and that $a+b+c=0$

So the numerator can be nicely simplified as $(0^2-2(ab+ac+bc))^2-2(ab+ac+bc)^2\\=4(ab+ac+bc)^2-2(ab+ac+bc)^2\\=2(ab+ac+bc)^2$

So $\frac{a^4+b^4+c^4}{a^2b^2+c^2(a^2+b^2)}=\frac{2(ab+ac+bc)^2}{(ab+ac+bc)^2}=\boxed{2}$

$(a+b+c)^2 = a^2 +b^2 +c^2 +2(ab +ac +bc) = 0$ $\Rightarrow\ (a^2 +b^2 +c^2)^2 = 4 (ab+ac+bc)^2$ $a^4 +b^4 +c^4 +2(a^2 b^2 +b^2 c^2 +a^2 c^2)=4(a^2 b^2 +a^2 c^2 +b^2 c^2)+8abc(a+b+c)$ From a+b+c = 0: $a^4 +b^4 +c^4 = 2 (a^2 b^2 +a^2 c^2 +b^2 c^2 )$ $\therefore\frac{a^4 +b^4 +c^4}{a^2 b^2 +c^2 (a^2 +b^2)} = \frac{2(a^2 b^2 +a^2 c^2 +b^2 c^2 )}{(a^2 b^2 +a^2 c^2 +b^2 c^2 )} = \boxed{2}$

(a^4+b^4+c^4)= (a^2+b^2+c^2)^2 - 2(a^2b^2+b^2c^2+c^2a^2) use this relation.