Question

If $\alpha$ and $\beta$ are the zeros of polynomial $x^2-ax+b$, then the value of ${\alpha }^2(\frac{{\alpha }^2}{\beta }-\beta )+{\beta }^2(\frac{{\beta }^2}{\alpha }-\alpha )$ is

• A. $\frac{a(a^2-4b)(a^2-b)}{b}$
• B. $\frac{b(a^2-4b)(a^2-b)}{a}$
• C. $\frac{b^2(a^2-4b)(a^2-b)}{a}$
• D. None

Answer 1

Answer: (A)

$\frac{a(a^2-4b)(a^2-b)}{b}$

We have Sum of the roots$=\alpha +\beta =a$

Product of the roots$=\alpha \beta =b$

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

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

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

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

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

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

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

We know that ${\alpha }^2+{\beta }^2={(\alpha +\beta )}^2-2\alpha \beta$ and

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

Using $\alpha +\beta =a$ and $\alpha \beta =b$ we have

${(\alpha -\beta )}^2={(\alpha -\beta )}^2=a^2-4b$

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

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

$=\frac{a(a^2-4b)(a^2-b)}{b}$