$\alpha$ and $\beta$ are the roots of the equation $x^2-a x+b=0$. If $\alpha^2+\beta^2$ and…
$\alpha$ and $\beta$ are the roots of the equation $x^2-a x+b=0$. If $\alpha^2+\beta^2$ and $\alpha^3+\beta^3$ are the roots of the equation $\mathrm{Ax}^2+\mathrm{Bx}+\mathrm{C}=0$, then $\mathrm{C}=$
$a^5-5 a^3 b+6 a b^2$
$a^5+5 a^3 b-6 a b^2$
$a^5-5 a^3 b-6 a b^2$
$a^5+5 a^3 b+6 a b^2$
Solution
Since $\alpha+\beta=a, \alpha \beta=b$
Now $\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta=a^2-2 b$
$\alpha^3+\beta^3=(\alpha+\beta)\left(\alpha^2+\beta^2-\alpha \beta\right)=a\left(a^2-3 b\right)$
So, quadratic equation
$\left(x-\left(a^2-2 b\right)\left(x-\left(a^3-3 a b\right)\right)\right.$
Thus constant term
$\begin{aligned}
& C=\left(a^2-2 b\right)\left(a^3-3 a b\right) \\
& =a^5-5 a^3 b+6 a b^2
\end{aligned}$