Question

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

Answer 1

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

We know that,

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

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

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

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

Now, $\sum (\frac{b}{c}+\frac{c}{b})=\frac{b+c+d}{a}+\frac{a+c+d}{b}+\frac{a+b+d}{c}+\frac{a+b+c}{d}$

$⟹\sum (\frac{b}{c}+\frac{c}{b})=\frac{\sum a^{2}bc}{abcd}\dots \dots \dots \dots .\left(5\right)$

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

$\sum (\frac{b}{c}+\frac{c}{b})=\frac{pr-4s}{s}$