<s>[INST] Surd Manipulations. If a=$-\sqrt {99}+\sqrt {999}+\sqrt {9999}$
b=$-\sqrt {99}-\sqrt {999}+\sqrt {9999}$
c=$-\sqrt {99}+\sqrt {999}-\sqrt {9999}$
Then $\displaystyle\sum_\limits{cyc} \frac{a^4}{(a-b)(a-c)}$ equals??
I first individually calculated (a-b), (b-c), (c-a) and tried to simplify the sum. However, I couldn't do it. Please help!
 [/INST] Hint:
$$a^4(b-c)+b^4(c-a)+c^4(a-b)$$
$$=ab(a^3-b^3)-c(a^4-b^4)+c^4(a-b)$$
$$=(a-b)\{ab(a^2+ab+b^2)-c(a+b)(a^2+b^2)+c^4\}$$
$$=\cdots$$
$$=-(a-b)(b-c)(c-a)(a^2+b^2+c^2+ab+bc+ca)$$
Alternatively, if $f(a,b,c)=a^4(b-c)+b^4(c-a)+c^4(a-b)$
Clearly, $f(a,a,c)=0\implies a-b$ is a factor
Similarly, $(b-c)(c-a)$ are also factors
As $f(a,b,c)$ is symmetric about $a,b,c$
$$f(a,b,c)=(a-b)(b-c)(c-a)(p(a^2+b^2+c^2)+q(ab+bc+ca))$$
Compare the coefficients of $a^4$ to find $p=-1$
Can you find $q=-1?$
Now $a^2+b^2+c^2+ab+bc+ca=(a+b+c)^2-3(ab+bc+ca)$
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