Sunaina thinks absurd

Well..  I was asked to prove Newton Sums, so since I latexed the proof, why not post it :P. Newton Sums: Consider a polynomial $P(x)$ of degree $n$, with roots $x_1,x_2,\dots,x_n$ $$P(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0.$$  Define $p_d=x_1^d+\dots+x_n^d.$ Then we have,   $$a_np_1 + a_{n-1} = 0$$  $$a_np_2 + a_{n-1}p_1 + 2a_{n-2}=0$$ $$a_np_3 + a_{n-1}p_2 + a_{n-2}p_1 + 3a_{n-3}=0$$ $$\vdots$$ (Define $a_j = 0$ for $j<0$.) Proof : Note that $$p_1=x_1+x_2+\dots+x_n=\frac{-a_{n-1}}{a_n}\implies \boxed{a_n\cdot p_1+a_{n-1}=0}$$ Note that $$p_2=x_1^2+\dots+x_n^2=(x_1+\dots+x_n)^2-2(x_1\cdot x_2+x_1\cdot x_3+\dots+x_{n-1}\cdot x_{n})=$$ $$p_1 \cdot \frac{-a_{n-1}}{a_n}-2\cdot \frac{a_{n-2}}{a_n}\implies \boxed{p_2\cdot a_n+p_1\cdot a_{n-1}+2\cdot a_{n-2}=0}$$ Note that $$p_3=x_1^3+\dots+x_n^3=(x_1^2+\dots+x_n^2)(x_1+\dots+x_n)-(x_1\cdot x_2+x_1\cdot x_3+\dots+$$ $$ x_{n-1}\cdot x_{n})(x_1+x_2+\dots+x_n)+3(x_1\cdot x_2\cdot x_3+\dots+x_{n-2}\cdot x_...

Well... I haven't seen much symmetric polynomials in Olympiads, but still I am learning, because I found them cute. And I am basically using this blog as my notes :P What are symmetric polynomials?  One can understand this with  examples. If we are considering over 3 variables, $x_1,x_2,x_3$ then  $$\sum_{sym}x_1^2\cdot x_2^3\cdot x_3=x_1^2\cdot x_2^3\cdot x_3+x_1^2\cdot x_3^3\cdot x_2+x_2^2\cdot x_1^3\cdot x_3+x_2^2\cdot x_3^3\cdot x_1+x_3^2\cdot x_1^3\cdot x_2.$$ See? $3!$ terms! Let's take one more example with again over 3 variables, $x_1,x_2,x_3$ then $$\sum_{sym}x_1^2\cdot x_2^2= x_1^2\cdot x_2^2+x_1^2\cdot x_3^2+x_2^2\cdot x_1^2+x_2^2\cdot x_3^2+x_3^2\cdot x_1^2+x_3^2\cdot x_2^2$$ Wait.. why 2 times ? So basically what happens in symmetrictric sums, is we go through all $n!$ possible permutations. So, here we have $a^2\cdot b^2\cdot c^0$ as like the "general" form type, right? Now, list down all the $3!=6$ permutations of $x_1,x_2,x_3$, and put them in the gene...