<s>[INST] Solving two Cubic Equation on their Roots. 
Let $x^{3}+ax+10=0$ and $x^{3}+bx^{2}+50=0$ have two roots in common. Let $P$ be the product of these common roots. Find the numerical value of $P^{3}$, not involving $a,b$.

My attempts:
Let roots of $x^{3}+ax+10=0$ be $\alpha,\beta,\gamma\implies$ 
\begin{align*}
&\alpha+\beta+\gamma =0 \\
&\alpha\beta+\alpha\gamma+\beta\gamma =a \\
&\alpha\beta\gamma = -10
\end{align*}
and of $x^{3}+bx^{2}+50=0$ be $\alpha,\beta,\gamma'\implies$
\begin{align*}
&\alpha+\beta+\gamma' =-b \\
&\alpha\beta+\alpha\gamma'+\beta\gamma' =0 \\
&\alpha\beta\gamma' = -50
\end{align*}
Few important equations:

  
*
  
*$\dfrac{\alpha\beta\gamma'}{\alpha\beta\gamma}=\dfrac{-50}{-10}=5$ $\implies \gamma'=5\gamma $
  
*$\gamma-\gamma'=b \because$ Substracting first eq
  
*$(\alpha+\beta)(\gamma-\gamma')=a$ Substracting second eq. from above
  
*$\alpha+\beta=\dfrac{a}{b} \ \ \because(2),(3)$ squaring gives: $(\alpha+\beta)^{2}=\dfrac{a^{2}}{b^{2}}$
  
*$\alpha\beta\gamma-\alpha\beta\gamma'=40\implies\alpha\beta(\gamma-\gamma')=40\implies\alpha\beta=\dfrac{40}{b}$
  

Also $(\alpha+\beta+\gamma')^{2} =b^{2}$
$\implies \alpha^2+\beta^2+\gamma'^2=b^2 \implies \alpha^2+\beta^2=b^2-\gamma'^2=b^2-\dfrac{25b^{2}}{16}=\dfrac{-9b^2}{16}\because (2),(1)$
Now, squaring first eq. of first set of eq.$\implies \alpha^2+\beta^2+\gamma^2+2a =0 \implies \alpha^2+\beta^2=\dfrac{-b^2}{16}-2a\because (2),(1)$
Equating this with previous equations, $\implies \alpha^2+\beta^2=\dfrac{-9b^2}{16}=\dfrac{-b^2}{16}-2a\implies a^2=\dfrac{b^4}{16}\rightarrow(7)$
Also, $\alpha^2 +\beta^2 +2\alpha\beta=\dfrac{a^2}{b^2} \because(4)$. And we calculated very thing in terms of $b$, putting all these,$\implies \dfrac{-9b^2}{16} +2\alpha\beta=\dfrac{b^4}{16b^2}\because(7)\implies \dfrac{80}{b}=\dfrac{10b^2}{16}\ \because (5)\implies b^3=128\rightarrow(6)$
Now, $(eq.5)^3\implies (\alpha\beta)^3=\dfrac{64000}{b^3}=\dfrac{64000}{128}=500=P^3\because(6)$
Is this correct, and if, then what are easiest/shortest method besides my GIANT method.
 [/INST] I thought that there was an impossibility that such polynomials exist. But in fact this is not the case, as evidenced by the edit @dxiv has done in his answer.
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