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.
 A: Good advice from George Polya in How to Solve It: keep the aim in mind.  It seems to me that you are writing down equations and then doing anything you can think of with them (and you have got the right answer).  But let's take a more structured view.
First, we don't want the answer to contain $a$ or $b$, so let's ignore them from the start.  Also, we want to find the product of roots, so let's write $P=\alpha\beta$.  It also appears from the equations that the sum of roots will be relevant, so let's write $S=\alpha+\beta$.  Then we have four equations
$$S+\gamma=0\ ,\quad P+S\gamma'=0\ ,\quad P\gamma=-10\ ,\quad P\gamma'=-50\ .$$
Call these equations $(1)$ to $(4)$.  Multiplying $(2)$ by $P$ and using $(4)$ gives
$$P^2-50S=0\ .$$
Using $(1)$ to eliminate $S$ gives
$$P^2=-50\gamma\ .$$
Multiplying by $P$ and using $(3)$, finally
$$P^3=500\ .$$
A: With OP's notations, the common roots $\alpha,\beta$ must satisfy the quadratic that results by subtracting the two cubics: $\,b x^2 - ax + 40 = 0\,$. Then by Vieta's formulas $\,P = \alpha \beta = \cfrac{40}{b}\,$.
Again by Vieta's, the 3rd root of the 2nd cubic is $\,\gamma' = -\cfrac{50}{\alpha \beta} = -\cfrac{5}{4} \,b\,$. Substituting in the equation:
$$
-\frac{125}{64} \,b^3 + b \cdot \frac{25}{16} \,b^2 + 50 = 0 \quad\iff\quad \cfrac{25}{64} \,b^3 = 50 \quad\iff\quad b^3 = 128
$$
Therefore $P^3 = \cfrac{40^3}{b^3}=\cfrac{512\cdot125}{128}=500\,$.

[ EDIT ] The following answers @JeanMarie's interesting question of whether such polynomials do in fact exist which satisfy the given condition.

The answer above proved that $b^3 = 128 \iff b=4\sqrt[3]{2}$. Then turning back to the 1st equation, its 3rd root must be $\gamma = -\cfrac{10}{P}=-\cfrac{1}{4}b=-\sqrt[3]{2}$. Substituting $\gamma$ into the 1st equation gives $a = 4 \sqrt[3]{4}$. Then the two equations are determined to be:
$$
\begin{cases}
\begin{align}
x^3 + 4 \sqrt[3]{4} \,x + 10 & = 0 \\
x^3 + 4 \sqrt[3]{2} \,x^2 + 50 & = 0
\end{align}
\end{cases}
$$
It is straightforward to verify that they do in fact have the quadratic $
x^2 - \sqrt[3]{2}\,x + 5 \sqrt[3]{4}$ as a common factor, so they have the two common roots $ (1 \pm i \sqrt{19})/\sqrt[3]{4}\,$. (Of course, their product is $P = 10 / \sqrt[3]{2}$ so this could be construed as another, painstaking way to answer OP's question).
A: 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.
