If $$a+b+c=0$$, then the value of $$\frac{(a^2+b^2+c^2)^2}{a^4+b^4+c^4}$$ is

Let $$S=a^2+b^2+c^2$$. Since $$a+b+c=0$$, 

$$(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$$

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

We have $$ab+bc+ca=-\dfrac{S}{2}$$ 

$$a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2=\dfrac{S^2}{4}$$. 

Now we also know that $$(a^2+b^2+c^2)^2 = a^4+b^4+c^4 +2(a^2b^2+b^2c^2+c^2a^2)$$

Thus $$a^4+b^4+c^4=S^2-2\left(\dfrac{S^2}{4}\right)=\dfrac{S^2}{2}$$

Hence the required ratio is $$ \dfrac{S^2}{\dfrac{S^2}{2}} = 2$$.