<s>[INST] $a+b=c+d$ , $a^3+b^3=c^3+d^3$ , prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$ 
Let $a, b, c, d$ be four numbers such that $a + b = c + d$ and $a^3 + b^3 = c^3 + d^3$. Prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$.

I've got $a^2+b^2-ab=c^2+d^2-cd$.
I tried squaring or cubing it repeatedly but I didn't get what I wanted. 
Now how do I proceed? 
 [/INST] $$(a+b)^3=(c+d)^3$$
$$\iff a^3+b^3+3ab(a+b)=c^3+d^3+3cd(c+d)$$
If $a+b\ne0, ab=cd\  \ \  \ (1)$
$\iff\dfrac ad=\dfrac cb=k$(say) $\  \ \  \ (2)$
$a+b=c+d\implies dk+b=bk+d\iff d(k-1)=b(k-1)$
Either $d=b\iff a=c$ 
or $k=1$ use this in $(2)$
Or Using $(1), ab=cda+b=c+d$ So if $a,b$ are roots of $t^2+bt+c=0,$ 
$c,d$ will be the roots of same equation $\implies\{a,b\}\equiv\{c,d\}$ 
In both bases  we can prove $$a^n+b^n=c^n+d^n$$
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