Question

If $\alpha$, $\beta$ are the roots of the equation $a{x}^2+bx+c=0$, the value of $\frac{{\alpha }^3}{\beta }+\frac{{\beta }^3}{\alpha }$ will be

• A. $\frac{b^4+2a^2{c}^2-4a{b}^{2}c}{a^{3}c}$
• B. $\frac{b^4+2a^2{c}^2-4a{b}^{2}c}{a^2{c}^2}$
• C. $\frac{b^4+2a^2{c}^2-4a{b}^{2}c}{a^4}$
• D. $\frac{3abc-b^3}{a{c}^2}$

Answer 1

The correct option is A $\frac{b^4+2a^2{c}^2-4a{b}^{2}c}{a^{3}c}$

If the roots are $\alpha$, $\beta$ of the equation

$a{x}^2+bx+c=0$

We know that,

$\alpha +\beta =-\frac{b}{a}$

and $\alpha \cdot \beta =\frac{c}{a}$

Given, $\frac{{\alpha }^3}{\beta }+\frac{{\beta }^3}{\alpha }=\frac{{\alpha }^4+{\beta }^4}{\alpha \cdot \beta }$

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

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

Putting the values,

$=\frac{{[\frac{b^2}{a^2}-\frac{2c}{a}]}^2-\frac{2c^2}{a^2}}{\frac{c}{a}}$

$=\frac{{\left[\frac{b^2-2ac}{a^2}\right]}^2-\frac{2c^2}{a^2}}{\frac{c}{a}}$

$=\frac{b^4+4a^2{c}^2-4a{b}^{2}c-2a^2{c}^2}{\left(\frac{a^4}{\left(\frac{c}{a}\right)}\right)}$

$=\frac{b^4+4a^2{c}^2-4a{b}^{2}c-2a^2{c}^2}{a^4}\times \frac{a}{c}$

$=\frac{b^4+2a^2{c}^2-4a{b}^{2}c}{a^{3}c}$