Question

If $a,b,c$ are the roots of the equation $x^3-p{x}^2+qx-r=0$, find the value of $\sum a^2{b}^2$.

Answer 1

Given, $x^4+p{x}^3+q{x}^4+rx+s=0$

We know that,

$\sum a=a+b+c+d=-p\dots \dots \dots \dots \dots .\left(1\right)$

$\sum ab=ab+ac+ad+bc+bd+cd=q\dots \dots \dots \dots \dots .\left(2\right)$

$\sum abc=abc+abd+acd+bcd=-r\dots \dots \dots \dots \dots .\left(3\right)$

$abcd=s\dots \dots \dots \dots \dots .\left(4\right)$

Squaring $\sum ab$, we get

$(\sum ab{)}^2=(ab+ac+ad+bc+bd+cd{)}^2$

$⟹(\sum ab{)}^2=\sum a^2{b}^2+2(a^{2}bc+a^{2}bd+a^{2}cd+a{b}^{2}c+a{b}^{2}d+b^{2}cd+ab{c}^2+a{c}^{2}d+b{c}^{2}d+ab{d}^2+ac{d}^2+bc{d}^2)+6abcd$

$⟹(\sum ab{)}^2=\sum a^2{b}^2+2\sum a^{2}bc+6abcd$

$⟹\sum a^2{b}^2=(\sum ab{)}^2-2\sum a^{2}bc-6abcd\dots \dots \dots \dots \dots .(5)$

Now, Multiplying $\sum a$ with $\sum abc$, we get

$(a+b+c+d)(abc+abd+acd+bcd)=\sum a^{2}bc+4abcd$

$⟹\sum a^{2}bc=\sum a\sum abc-4abcd=(-p)\cdot (-r)–4s=pr–4s$

Substituting the value of $\sum a^{2}bc$ in (5), we get

$\sum a^2{b}^2=q^2-2(pr-4s)–6s=q^2–2pr+8s-6s=q^2–2pr+2s$