Question

If $a,b,c$ are the roots of the equation $x^4-p{x}^2+qx-r=0$, find the value of $(b+c)(c+a)(a+b)$.

Answer 1

Given equation, $x^4+p{x}^3+q{x}^2+rx+s=0\dots \dots \dots \dots \dots \dots \left(1\right)$

Let $a,b,c,d$ be the roots of the given equation

$(b+c)(c+a)(a+b)=a^{2}b+a^{2}c+a{b}^2+b^{2}c+a{c}^2+b{c}^2+2abc$

$\therefore (b+c)(c+a)(a+b)+(b+d)(d+a)(a+b)+(c+d)(d+a)(a+c)+(c+d)(d+b)(b+c)$

$=2\sum a^{2}b+2\sum abc$

$=2(\sum a\sum ab-3\sum abc)+2\sum abc$

$=2\sum a\sum ab-4\sum abc$

$=-2pq+4r$