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Consider $$x^3 + px + q = 0$$
with roots $ alpha, beta. gamma$

We want to find the degree 3 polynomial that has roots :
$$ fracalphabetagamma, frac alphagammabeta, frac betagamma alpha $$

My attempt so far:

$$frac alphabetagamma = fracalphabetagammagamma^2 = - fracqgamma^2 $$
and similarly for all of the others:

So now I let $$ t = - fracqx^2 $$ and substitute into the original polynomial; except I will multiply everything by $x$ first to make my substitution easier.

$$ x^4 + px^2 = -qx $$
$$ fracq^2t^2 - fracpqt = - qsqrt -fracqt $$
Squaring both sides and multiplying by $t^4$, we obtain:
$$qt^3 +p^2t^2 -2pqt +q^2 = 0 $$

Except this seems like a messy, tedious solution as you have to muck around with square roots etc. Is there a nicer way to do this?

Here is a hint for a way of solving it using elementary steps: Consider a degree 3 polynomial that have roots $alpha, beta, gamma$:

$$
(x-alpha)(x-beta)(x-gamma),
$$
this must then be the same polynomial as
$$
x^3+px+q.
$$

Can you now relate now $alpha, beta, gamma$ with $p$ and $q$? Do we get any relation between $alpha, beta, gamma$?

Yes, from
$$
(x-alpha)(x-beta)(x-gamma)=x^3+px+q
$$
one obtains $alpha beta + alpha gamma + betagamma= p$, $alpha beta gamma =-q$, and $alpha + beta + gamma=0$. Due to this latter relation, one can also write $(alpha + beta + gamma)^2=0$, i.e., $alpha^2 +beta^2 +gamma^2=-2(alpha beta + alpha gamma + betagamma)=-2p$ and also $(alpha beta)^2 + (alpha gamma)^2 + (beta gamma)^2=(alpha beta + alpha gamma + beta gamma)^2-2(alpha betagamma)(alpha +beta+gamma)=(alpha beta + alpha gamma + beta gamma)^2=p^2$

After this, consider a degree 3 function that has roots $fracalpha betagamma, fracalpha gammabeta, fracalpha gammabeta$:

$$
left(x-fracalpha betagammaright)left(x-fracalpha gammabetaright)left(x-fracalpha gammabetaright)=\
frac1alpha beta gamma
left[(alpha beta gamma) x^3 - ((alpha beta)^2 + (alpha gamma)^2+(beta gamma)^2)x^2 + (alpha beta gamma)(alpha^2 + beta^2 +gamma^2)x -(alphabetagamma)^2right]
$$

Now use the obtained identities between $alpha, beta, gamma$ and $p, q$ to write up the above degree 3 polynomial with the desired roots:

$$
qx^3+p^2x^2-2pqx +q^2.
$$

You just have to express the elementary symmetric functions of $;dfracalphabetagamma, dfracbetagammaalpha$ and $dfracgammaalphabeta$ with $p, q,r$, then use Vieta's relations.

• $beginaligned[t]dfracalphabetagamma+ dfracbetagammaalpha+dfracgammaalphabeta&=dfrac(alphabeta)^2+(betagamma)^2+(gammaalpha)^2alphabetagamma=dfrac(alphabeta+betagamma+gammaalpha)^2-2(alpha^2betagamma+alphabeta^2gamma+alphabetagamma^2)alphabetagamma\&=dfrac(alphabeta+betagamma+gammaalpha)^2alphabetagamma-2(alpha+beta+gamma)=fracp^2-rendaligned
  $


• $dfracalphabetagammacdotdfracbetagammaalpha+dfracbetagammaalphacdotdfracgammaalphabeta+dfracgammaalphabetacdotdfracalphabetagamma=alpha^2+beta^2+gamma^2=(alpha+beta+gamma)^2-2(alphabeta+betagamma+gammaalpha)=-2p$


• $dfracalphabetagammacdotdfracbetagammaalphacdotdfracgammaalphabeta=alphabetagamma=-r$

Hence the equation is
$$x^3+fracp^2r x^2-2px+r=0iff rx^3+p^2x^2-2prx+r^2=0.$$