Pls keep answer with solution

1.	Pls keep answer with solution
Answer» $\displaystyle\red{\underline{\underline{Solution}}}$ LET $\displaystyle \: \: \alpha \: \: and \: \: { \alpha}^{2}$ are the ROOTS of the equation $\displaystyle \: \: {x}^{2} + px + q = 0$ Then $\displaystyle \: \: \alpha \: + { \alpha}^{2} = - p \: \: \: \: and \: \: \displaystyle \: \: \alpha \: \times \: { \alpha}^{2} = q$ Which gives $\displaystyle \: \: \alpha (<klux>1</klux> + { \alpha}) = - p \: \: \: \: .........(1)$ $\displaystyle \: \: \: { \alpha}^{3} = q \: \: \: ..........(2)$ Now Taking CUBE in both sides of (1) $\displaystyle \: \: {p}^{3}$ $= - \displaystyle \: \: { \alpha}^{3} ({1 \: \: + \: \: { \alpha})}^{3}$ $= - \displaystyle \: \: { \alpha}^{3} \{{1 \: \: + \: \: { \alpha}}^{3} + 3 { \alpha} {(1 + \alpha})\}$ $\displaystyle \: = - q(1 + q - 3p) \: \: by \: \: (1) \: \: \: and \: \: \: (2)$ $\displaystyle \: = - q - {q}^{2} + 3pq$ So $\displaystyle \: {p}^{3} + {q}^{2} = - q + 3pq$ Hence $\displaystyle \: {p}^{3} + {q}^{2} = q( 3p - 1)$

Answer»

$\displaystyle\red{\underline{\underline{Solution}}}$

LET $\displaystyle \: \: \alpha \: \: and \: \: { \alpha}^{2}$ are the ROOTS of the equation $\displaystyle \: \: {x}^{2} + px + q = 0$