If $\alpha, \beta$ are the roots of $x^2+p x+q=0$, then the values of $\alpha^3+\beta^3$ and…
If $\alpha, \beta$ are the roots of $x^2+p x+q=0$, then the values of $\alpha^3+\beta^3$ and $\alpha^4+\alpha^2 \beta^2+\beta^4$ are respectively ...... and ......
$\left(3 p q-p^3\right)$ and $\left(p^4-3 p^2 q+3 q^2\right)$
$-p\left(3 q-p^2\right)$ and $\left(p^2-q\right)\left(p^2+3 q\right)$
$(p q-4)$ and $\left(p^4-q^4\right)$
$\left(3 p q-p^3\right)$ and $\left(p^2-q\right)\left(p^2-3 q\right)$
Solution
Since $\alpha$ and $\beta$ are roots of equation
$
\begin{aligned}
& x^2+p x+q=0, \text { so } \\
& \alpha+\beta=-p \text { and } \alpha \beta=q
\end{aligned}
$
As we know that, $\alpha^3+\beta^3=(\alpha+\beta)\left(\alpha^2+\beta^2-\alpha \beta\right)$
$
\begin{aligned}
& =(\alpha+\beta)\left[(\alpha+\beta)^2-3 \alpha \beta\right] \\
& =(-p)\left[p^2-3 q\right]=3 p q-p^3 \\
\text { and } & \alpha^4+\alpha^2 \beta^2+\beta^4 \\
& =\left(\alpha^2+\beta^2\right)^2-\alpha^2 \beta^2=\left[(\alpha+\beta)^2-2 \alpha \beta\right]^2-(\alpha \beta)^2 \\
& =\left((-p)^2-2 q\right)^2-q^2 \\
& =\left(p^2-2 q\right)^2-q^2=\left(p^2-2 q-q\right)\left(p^2-2 q+q\right) \\
& =\left(p^2-q\right)\left(p^2-3 q\right)
\end{aligned}
$