Finding Polynomial

Let $p$ be a root of the equation $P(x)=x^3-3x+1=0$

Then, $p^3-3p=-1$ or, $-1=p(p^2-3)$

Raising both sides to the 5th power, we get

$-1=p^5(p^2-3)^5=p^5(p^{10}-15p^8+90p^6-270p^4+405p^2-243)$

or, $-1=p^5(p^{10}-243-15(p^4-3p^2)(p(p^3-3p)+9))$

Since $p^3-3p=-1$ , we get

$-1=p^5(p^{10}-243+15(p^5-3p^3-9p(p^3-3p)))$

or $-1=p^5(p^{10}-243+15(p^5-3(p^3-3p)))$

or, $-1=p^5(p^{10}-243+15(p^5+3))$

Therefore, $p^{15}+15p^{10}-198p^5+1=0$

Let $q=p^5$ be a root of the polynomial equation $Q(x)=0$ . Then, $q^3+15q^2-198q+1=0$

Therefore, $Q(x)=x^3+15x^2-198x+1$

Let $\alpha ,\beta ,\gamma$ be the roots of $P(x)$ .

So

$\alpha \beta \gamma =-1$

$\alpha +\beta +\gamma =0$

$\alpha \beta +\beta \gamma +\gamma \alpha =-3$

Now as $\alpha$ is a root of $P(x)$

so ${ \alpha }^{ 3 }-3\alpha +1=0$

On Multiplying this eq by $\alpha^{2}$ we get

${ \alpha }^{ 5 }-{ 3\alpha }^{ 3 }+{ \alpha }^{ 2 }=0$

Similarly we get ${ \gamma }^{ 5 }-{ 3\gamma }^{ 3 }+{ \gamma }^{ 2 }=0$

and ${ \beta }^{ 5 }-{ 3\beta }^{ 3 }+{ \beta }^{ 2 }=0$

On adding these equations we get

${ \alpha }^{ 5 }+{ \beta }^{ 5 }+{ \gamma }^{ 5 }=3({ \alpha }^{ 3 }+{ \beta }^{ 3 }+{ \gamma }^{ 3 })-({ \alpha }^{ 2 }+{ \beta }^{ 2 }+{ \gamma }^{ 2 })$

Using algebraic identities with starting equations we get

${ \alpha }^{ 5 }+{ \beta }^{ 5 }+{ \gamma }^{ 5 }=-15$

and ${ (\alpha \beta \gamma ) }^{ 5 }=-1$

As the polynomial which is formed is of 3 degree whose roots are the fifth powers of the roots of $P(x)$

So ${ \alpha }^{ 5 }+{ \beta }^{ 5 }+{ \gamma }^{ 5 }=-\frac { { a }_{ 1 } }{ { a }_{ 0 } }$

and ${ (\alpha \beta \gamma ) }^{ 5 }=-\frac { { a }_{ 3 } }{ { a }_{ 0 } }$

Here $a_{0}, a_{1}, a_{3}$ are the coefficients of $x^{3}$ and $x^{2}$ and constant respectively.

So $a_{1}=15$ and $a_{3}=1$ . So the answer is ${ x }^{ 3 }+{ 15x }^{ 2 }-198x+1$

Here I don't bother to find the coefficient of $x$ because in every option there is same coefficient.