If $x^3+px+r$ and $3x^2+p$ have a common factor, prove that $\dfrac{p^3}{27}+\dfrac{r^2}{4}=0$

Question

If $x^3+px+r$ and $3x^2+p$ have a common factor, prove that $\dfrac{p^3}{27}+\dfrac{r^2}{4}=0$

So if there is a common factor, let the common root be $t$, and then $t^3+pt+r=3t^2+p$

$\Rightarrow t^3-3t^2+pt=p-r$

Then I'm stuck. I don't know how to derive the value of $t$ in terms of $p$ and $r$ in this third-order equation. Thanks for the help.

Answer 1

HINT

Eliminate $t^3$ from $t^3+pt+r=0$ and $3t^2+p=0$. (First multiply the first equation by $3$ and multiply the second equation by $t$.)

You will obtain a simple expression for $t$ and then substitute this into $3t^2+p=0$.

Answer 2

Another approach is to apply the Viete relations. The zeroes of $ \ 3x^2 + p \ $ are given by $ \ \alpha^2 \ = \ -\frac{p}{3} $ $ \Rightarrow \ -3 \alpha^2 \ = \ p \ \Rightarrow \ p^3 \ = \ -27·\alpha^6 \ \ . \ $ In the cubic polynomial $ \ x^3 + px + r \ \ , \ $ the zeroes are $ \ \pm \alpha \ , \ \beta \ , \ \gamma \ \ , \ $ the first of these has one sign or the other (it won't matter) and the other two zeroes may be real or a complex-conjugate pair (that also won't matter). We then have $$ p \ \ = \ \ \alpha \beta \ + \ \alpha \gamma \ + \ \beta \gamma \ \ = \ \ \alpha·(\beta \ + \ \gamma) \ + \ \beta \gamma \ \ . \ $$ Since we also have $ \ \alpha + \beta + \gamma \ = \ 0 \ \ $ and $ \ \alpha · \beta · \gamma \ = \ -r \ \ , \ $ it follows that $$ p \ \ = \ \ \alpha·(-\alpha) \ + \ \frac{-r}{\alpha} \ \ \Rightarrow \ \ \alpha·(-3·\alpha^2) \ \ = \ \ -\alpha^3 \ - \ r \ \ \Rightarrow \ \ r \ \ = \ \ 2·\alpha^3 \ \ . \ $$ Consequently, $$ \alpha^6 \ \ = \ \ -\frac{p^3}{27} \ \ = \ \ \left( \frac{r}{2} \right)^2 \ \ . $$