Question

pls keep answer with solution​

Answer 1

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

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

$\displaystyle \: \: \alpha (1 + { \alpha}) = - p \: \: \: \: .........(1)$

$\displaystyle \: \: \: { \alpha}^{3} = q \: \: \: ..........(2)$

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 2

Answer:

Let \displaystyle \: \: \alpha \: \: and \: \: { \alpha}^{2}αandα2 are the roots of the equation \displaystyle \: \: {x}^{2} + px + q = 0x2+px+q=0

Then

\displaystyle \: \: \alpha \: + { \alpha}^{2} = - p \: \: \: \: and \: \: \displaystyle \: \: \alpha \: \times \: { \alpha}^{2} = qα+α2=−pandα×α2=q

Which gives

\displaystyle \: \: \alpha (1 + { \alpha}) = - p \: \: \: \: .........(1)α(1+α)=−p.........(1)

\displaystyle \: \: \: { \alpha}^{3} = q \: \: \: ..........(2)α3=q..........(2)

Now

Taking Cube in both sides of (1)

\displaystyle \: \: {p}^{3}p3

= - \displaystyle \: \: { \alpha}^{3} ({1 \: \: + \: \: { \alpha})}^{3}=−α3(1+α)3

= - \displaystyle \: \: { \alpha}^{3} \{{1 \: \: + \: \: { \alpha}}^{3} + 3 { \alpha} {(1 + \alpha})\}=−α3{1+α3+3α(1+α)}

\displaystyle \: = - q(1 + q - 3p) \: \: by \: \: (1) \: \: \: and \: \: \: (2)=−q(1+q−3p)by(1)and(2)

\displaystyle \: = - q - {q}^{2} + 3pq=−q−q2+3pq

So

\displaystyle \: {p}^{3} + {q}^{2} = - q + 3pqp3+q2=−q+3pq

Hence

\displaystyle \: {p}^{3} + {q}^{2} = q( 3p - 1)p3+q2=q(3p−1)