The way I'm about to show you is used in a lot of questions like this, and pretty much always works, you just need to find the right manipulation of squaring and it will all come out in the end.
I can't type up all the working such as squaring something with 3 terms in it, but I've done it on paper and it is correct
1. (a^2 + b^2 +c^2)^2= a^4 + b^4+ c^4 + 2(a^2b^2 + c^2a^2 + b^2c^2)
2. (a+b+c)^2 = a^2 + b^2 +c^2 +2(ab+bc+ca)
3. (ab+bc+ca)^2 = a^2b^2 + c^2a^2 + b^2c^2 +2abc(a+b+c)
These 3 facts are the groundwork (and took me a while to work out, but note the pleasing symmetry to them all) and will allow you to, after a bit of "changing the subject", to equate to your initial conditions a+b+c=0 and a^2+b^2+c^2=1
By the way, to do something like (a+b+c)^2, the quickest method is to write it as (a+(b+c))^2 and do it as though a is one term and b+c is the other term
OKAY. Onwards.
Using 3. we can see that (ab+bc+ca)^2 - (a^2b^2+c^2a^2+b^2c^2) = 2abc(a+b+c) = 0 because a+b+c = 0
so we have this
4. (ab+bc+ca)^2 = (a^2b^2+c^2a^2+b^2c^2)
Using 1. we rewrite it as
5. (a^2+b^2+c^2) = a^4+b^4+c^4 +2(ab+bc+ca)^2
Now rewrite 5, using 2, as
6. (a^2+b^2+c^2)^2 = a^4+b^4+c^4 +2 [[ (a+b+c)^2 -(a^2+b^2+c^2)]/2]^2
Now as a^2+b^2+c^2=1 and a+b+c=0, 6. obviously can transform to
7. (1)^2 = a^4+b^4+c^4 +2[[0-1]/2]^2
1=a^4+b^4+c^4 +2[(-1/2)]^2
1=a^4+b^4+c^4 +2[1/4]
a^4+b^4+c^4 = 1/2
put this into 16/a^4+b^4+c^4 and you get 32.
For these kind of questions, simply finding a value which will fit won't work every time. It's better to understand a method like mine, or the other algebraic method someone's written here, which can be used in every case.
I think this method is the best, I've seen it used in more advanced cases such as relationships between x^n+y^n+z^n and the same for n+1 and n+2. And it's fairly simple, the first 3 steps are just casual expanding brackets, the next 3 are just moving things around and plugging in values! Much simpler than finding an X,Y and getting mixed up with too many variables.
Anyway, hope it helps :) x