Question

For $a,b\in R-\{0\}$, let $\alpha ,\beta$ be roots of the equation $x^2-ax+b=0$. The quadratic equation whose roots are $(\alpha -a{)}^{-3}$ and $(\beta -a{)}^{-3}$ is

• A. $b^3{x}^2+\left(a^3-3ab\right)x+1=0$
• B. $a^3{x}^2+\left(b^3-3ab\right)x+1=0$
• C. $b^3{x}^2+\left(a^3-ab\right)+1=0$
• D. $a^3{x}^2+\left(b^3-ab\right)x+1=0$

Answer 1

Answer: (A)

Now, ${\alpha }^2-a\alpha +b=0\Rightarrow \alpha (\alpha -a)=-b\Rightarrow (\alpha -a)=\frac{-b}{\alpha }$
Also, ${\beta }^2-a\beta +b=0\Rightarrow \beta (\beta -a)=-b\Rightarrow (\beta -a)=\frac{-b}{\beta }$
Now, sum of roots $=(\alpha -a{)}^{-3}+(\beta -a{)}^{-3}$
$=\frac{1}{(\alpha -a{)}^3}+\frac{1}{(\beta -a{)}^3}=\frac{-{\alpha }^3}{b^3}-\frac{{\beta }^3}{b^3}=\frac{-1}{b^3}\left[(\alpha +\beta {)}^3-3\alpha \beta (\alpha +\beta )\right]=\frac{-1}{b^3}\left(a^3-3ba\right)$
Also, product of roots $=(\alpha -a{)}^{-3}\cdot (\beta -a{)}^3$
$=\frac{1}{(\alpha -a{)}^3}\frac{1}{(\beta -a{)}^3}=\left(\frac{-{\alpha }^3}{b^3}\right)\left(\frac{-{\beta }^3}{b^3}\right)=\frac{{\alpha }^3{\beta }^3}{b^6}=\frac{b^3}{b^6}=\frac{1}{b^3}$
$\therefore$ The required quadratic equation is, $x^2-Sx+P=0$
$\Rightarrow x^2+\frac{1}{b^3}\left(a^3-3ab\right)x+\frac{1}{b^3}=0$
$\Rightarrow b^3{x}^2+\left(a^3-3ab\right)x+1=0$