# What's the best way to compute $\frac{a^4 + b^4 + c^4}{a^2 + b^2 + c^2}$

So, my teacher gave us this to compute yesterday, and I'm completly confused on how should I proceed :

$$\frac{1^4 + 2012^4 +2013^4}{1^2 + 2012^2 + 2013^2}$$

I've tried several ways, but most of them are very long, for example I've simplified both numbers :

$2012^2 = (2 * 10^3)^2 + 12^2 + 24 * 10^3$

$2013^2 = (2 * 10^3)^2 + 13^2 + 26 * 10^3$

I can't see how this could help me solve this problem . So how how should I go with this kind of problems in general ?

• For literal expresions nothing to do. But for numerical ones certainly you can act different ways according to the involved numbers. May 6, 2016 at 16:07

What you are looking at is $$\frac{1+x^4+(x+1)^4}{1+x^2+(x+1)^2}$$ where $x=2012$. Expanding and simplifying, this is $$\frac{2+4x+6x^2+4x^3+2x^4}{2+2x+2x^2}=\frac{1+2x+3x^2+2x^3+x^4}{1+x+x^2}$$ $$=\frac{(1+x+x^2)^2}{1+x+x^2}=1+x+x^2.$$

Thus the final answer is $$2012^2+2012+1,$$ and from here you can calculate the number by hand.

• Great Answer +1 May 6, 2016 at 12:06
• Why don't you accept if it great? This is as good as it gets. May 6, 2016 at 12:07

Hint The given triple $(1, 2012, 2013)$ has the special feature that $1 + 2012 = 2013$. This motivates writing the quotient as $$\frac{a^4 + b^4 + (a + b)^4}{a^2 + b^2 + (a + b)^2}$$ for $a = 1, b = 2012$.

Additional hint Expanding the numerator, we can see that it factors as $2(a^2+ab+b^2)^2$

• Good observation, but the answer by Eric, is more straightforward in my opinion . May 6, 2016 at 12:07
• Of course, the answers are essentially the same, except that Eric's specializes to the case $a = 1$. May 6, 2016 at 12:16

Hint:

$$\frac{1^4+x^4+(x+1)^4}{1^2+x^2+(x+1)^2}=x^2+x+1$$

• I'd love if you could devlope your answer with a proof of what you've said . May 6, 2016 at 12:04