Question

If $\alpha ,\beta$ are the roots of the equation $x^2-px+q=0$, find the values of
(1) ${\alpha }^2({\alpha }^2{\beta }^{-1}-\beta )+{\beta }^2({\beta }^2{\alpha }^{-1}-\alpha )$
(2) ${(\alpha -p)}^{-4}+{(\beta -p)}^{-4}$

Answer 1

Given, $\alpha ,\beta$ are roots of the equation $x^2-px+q=0$

$\therefore \alpha +\beta =p\text{and}\alpha \beta =q$

Part 1:

${\alpha }^2(\frac{{\alpha }^2}{\beta }-\beta )+{\beta }^2(\frac{{\beta }^2}{\alpha }-\alpha )={\alpha }^2\left(\frac{{\alpha }^2-{\beta }^2}{\beta }\right)+{\beta }^2\left(\frac{{\beta }^2-{\alpha }^2}{\alpha }\right)$

$=({\alpha }^2-{\beta }^2)(\frac{{\alpha }^2}{\beta }-\frac{{\beta }^2}{\alpha })=\frac{({\alpha }^2-{\beta }^2)({\alpha }^3-{\beta }^3)}{\alpha \beta }$

$=\frac{(\alpha +\beta )(\alpha -\beta {)}^2({\alpha }^2+{\beta }^2+\alpha \beta )}{\alpha \beta }$

$=\frac{(\alpha +\beta )\left[\right(\alpha +\beta {)}^2-4\alpha \beta \left]\right[(\alpha +\beta {)}^2-\alpha \beta ]}{\alpha \beta }$

$=\frac{p(p^2-4q)(p^2-q)}{q}$

Part 2:

$\frac{1}{(\alpha -p{)}^4}+\frac{1}{(\beta -p{)}^4}=\frac{1}{{\beta }^4}+\frac{1}{{\alpha }^4}=\frac{{\alpha }^4+{\beta }^4}{{\alpha }^4{\beta }^4}$

$=\frac{({\alpha }^2+{\beta }^2{)}^2-2{\alpha }^2{\beta }^2}{{\alpha }^4{\beta }^4}==\frac{\left(\right(\alpha +\beta {)}^2-2\alpha \beta {)}^2-2{\alpha }^2{\beta }^2}{{\alpha }^4{\beta }^4}$

$=\frac{(p^2-2q{)}^2-2q^2}{q^4}=\frac{p^2-4p^{2}q+2q^2}{q^4}$