Exercise

If \( \alpha, \beta\) are the roots of the equation \(ax^2+bx+c=0\), find the equation whose roots are \(\alpha ^2 \beta ^{-1}\)and \(\beta ^2 \alpha ^{-1}\)
Given quadratic equation is
\(ax^2+bx+c=0\)
Let \( \alpha,\beta\) are the roots of \(ax^2+bx+c=0\), then
\( \alpha+ \beta=-\frac{b}{a}, \alpha \beta=\frac{c}{a}\)
According to the question, the new roots are
\(\alpha ^2 \beta ^{-1}\) and \(\beta ^2 \alpha ^{-1}\)
So,

Sum of new roots	=\(\alpha ^2 \beta ^{-1}+ \beta ^2 \alpha ^{-1}\)
	=\( \frac{\alpha ^2}{\beta}+ \frac{\beta ^2}{\alpha}\)
	=\( \frac{\alpha ^3+\beta ^3}{\alpha \beta}\) =\( \frac{(\alpha +\beta) ^3-3\alpha \beta (\alpha +\beta) }{\alpha \beta}\) =\( \frac{(\alpha +\beta) ^3} {\alpha \beta} - \frac{3\alpha \beta (\alpha +\beta) }{\alpha \beta}\) =\( \frac{(\alpha +\beta) ^3} {\alpha \beta} - 3 (\alpha +\beta) \) =\( \frac{\left (-\frac{b}{a} \right ) ^3} {\frac{c}{a}} - 3 \left (-\frac{b}{a} \right ) \) =\( \frac{-b^3}{ca^2} + \frac{3b}{a} \) =\( \frac{3abc-b^3}{a^2c} \)

Again,

Product of new roots	=\(\alpha ^2 \beta ^{-1} . \beta ^2 \alpha ^{-1}\)
	=\( \frac{\alpha ^2}{\beta}. \frac{\beta ^2}{\alpha}\)
	=\( \alpha \beta\) =\( \frac{c}{a}\)

Now
The new quadratic equation is
\( x^2 - (\text {sum of roots}) x + (\text {product of roots}) =0\)
or \( x^2- \left ( \frac{3abc-b^3}{a^2c} \right) x + \left ( \frac{c}{a} \right)=0\)
or \( a^2cx^2+ (b^3-3abc)x + ac^2=0\)
This is the required quadratic equation