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$\ L $ is the tangent line at $\ x = x_0 $ to the graph of $\ y= ax^3 +bx$. I need to find the $\ x$ value of the second point where $\ L$ cuts through the graph.

I tried to define $\ (x_0, ax_0^3+bx_0) $ and then just find the equation of the tangent line at $\ x_0 $ which is $\ y = 3ax_0^2 -2ax_0^3 + bx $ and then just compare it to the graph so I get $\ 3ax_0^2-2ax_0^3 +bx = ax^3 +bx $ but all I get is $\ x^3 = 3x_0^2x-2x_0^3 $ and I can not get much out of it.

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\begin{eqnarray} x^3-3x_0^2x+2x_0^3 &=&x^3-x_0^2x-2x_0^2x+2x_0^3\\ &=& x(x^2-x_0^2)-2x_0^2(x-x_0)\\ & =& (x-x_0)\Big(x(x+x_0)-2x_0^2\Big)\\ &=&(x-x_0)\Big(x^2+xx_0-2x_0^2\Big)\\ &=&(x-x_0)(x+2x_0)(x-x_0)\\ &=&(x-x_0)^2(x+2x_0) \end{eqnarray} So it intersect the graph at $(-2x_0,...)$.

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You have to solve the equation $4x^{3}-3x_0^{2}x-2x_0^{3}=0$ and you already know that $x=x_0$ Is one solution. The equation can be written as $(x-x_0)^{2}(x+2x_0)=0$. Hence $x=-2x_0$is the answer.

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