If $a,b,c,d$ are the roots of the equation biquadratic equation $x^4+px^3 +qx^2+rx+s=0$ , find the value of $\Sigma a^2b^2$. If $a,b,c,d$ are the roots of the equation biquadratic equation $x^4+px^3 +qx^2+rx+s=0$
, find the value of $\Sigma a^2b^2$.
My solution goes like this:

Since, $a,b,c,d$ are the roots of the equation biquadratic equation $x^4+px^3+qx^2+rx+s=0$ so, we have, $\Sigma a=-p$,$\Sigma ab=q$,$\Sigma abc=-r$,$ abcd=s$. Now.$$\Sigma a^2b^2=\Sigma (ab)^2-2\Sigma a^2bc-6abcd.$$Also, $$\Sigma a^2bc=\Sigma a\Sigma abc -4abcd=pr-4s$$. Thus, $\Sigma a^2b^2=\Sigma (ab)^2-2\Sigma a^2bc-6abcd=q^2-2pr+2s.$

However, in this solution when I wrote the expression $\Sigma a^2b^2=\Sigma (ab)^2-2\Sigma a^2bc-6abcd$, I had to calculate the value of $(ab+ac+...+cd)^2$ manually and then I could write in this "above -short form"(which looks simplified just because of the notations used). Same goes for the calculation of $$\Sigma a^2bc=\Sigma a\Sigma abc -4abcd=pr-4s$$ . There , I had to calculate $(a+b+c+d)(abc+...+bcd)$ manually . However, these manual calculations are unusually large as well as huge . Is there any other way to directly predict $$\Sigma a^2bc=\Sigma a\Sigma abc -4abcd=pr-4s$$ and $\Sigma a^2b^2=\Sigma (ab)^2-2\Sigma a^2bc-6abcd$? Or is direct calculation the only way to arrive at these results?
 A: Symmetric functions are
an interesting and actively explored field of study. Suppose
we have a finite but arbitrary set of variables
$\{a,b,c,\dots\}.$ Define
some elementary symmetric functions
$$ e_1 \!=\! \Sigma a \!=\! a\!+\!b\!+\!c\!+\!\dots ,\;
 e_2 \!=\! \Sigma ab \!=\! ab\!+ac\!+\!bc\!+\!\dots, \;
e_3 \!=\! \Sigma abc \!=\! abc\!+\!\dots. $$
A natural question is how to find $\Sigma a^2.$ First, compute
$$ e_1^2 = (\Sigma a)(\Sigma a) = \Sigma aa+ \Sigma ab +\Sigma ba =
\Sigma a^2 + 2e_2 $$
which is a generalization of $(a+b)^2 = (a^2+b^2) + 2(ab)$ to
many variables. The result is
$\Sigma a^2 = e_1^2 - 2e_2.$ Similar results can be found.
For example,
$$ e_1e_2 = (\Sigma a)(\Sigma ab) = \Sigma a^2b +\Sigma bab
+3\Sigma abc = \Sigma a^2b + 3e_3. $$ The result is
$\Sigma a^2b = e_1e_2 - 3e_3.$ You have to be careful to count
the monomials with proper multiplicities. Another example is
$$ e_1e_3 = (\Sigma a)(\Sigma abc) = \Sigma a^2bc + 
\Sigma babc +\Sigma cabc+ 4e_4 = \Sigma a^2bc +4e_4 $$ which
generalizes. Final example is
$$ e_2e_2 = (\Sigma ab)(\Sigma ab) = \Sigma a^2b^2 +
 2\Sigma a^2bc + 6\Sigma abcd = \Sigma a^2b^2 + 2\Sigma a^2bc
+ 6e_4. $$ Combine with previous results
to get $\Sigma a^2b^2 = e_2^2 -2e_3e_1 +2e_4.$
You can check this result in the special case where all of
the variables are equal.
