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.

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).
