Question

If $a,b,c$ are the roots of $x^3+qx+r=0$, find the value of
$\left(1\right)(b-c{)}^2+(c-a{)}^2+(a-b{)}^2$.
$\left(2\right)(b+c{)}^{-1}+(c+a{)}^{-1}+(a+b{)}^{-1}$.

Answer 1

$x^3+qx+r=0$

Let the roots be $a,b,c$

$a+b+c=0ab+bc+ca=qabc=-r$

$-------------------------------------$

${(a-b)}^2+{(b-c)}^2+{(c-a)}^2=2(a^2+b^2+c^2)-2(ab+bc+ac)=2\{{(a+b+c)}^2-2(ab+bc+ac\left)\right\}-2(ab+bc+ac)=2{(a+b+c)}^2-6(ab+bc+ac)=0-6q=-6q$

$-------------------------------------$

${(b+c)}^{-1}+{(c+a)}^{-1}+{(a+b)}^{-1}=\sum \frac{1}{b+c}$

As $a+b+c=0$

$\Rightarrow \sum \frac{1}{b+c}=-\sum \frac{1}{a}=-\frac{ab+bc+ac}{abc}=-\frac{q}{-r}=\frac{q}{r}$