Solve the equation for $X$

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$$X^3-3X^2+3X=\frac{3R-10}{2}$$

How can i solve it for $X$ ?

I tried to do :

$$\Rightarrow X(X^2-3X+3)=\frac{3R-10}{2}$$ ???

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You have $x^3-3x^2+3x = c$. You can 'complete the cube' on the left hand side by noting that $x^3-3x^2+3x-1 = (x-1)^3$. To achieve this, subtract $1$ from both sides, $$x^3-3x^2+3x -1= c-1.$$

You should be able to use this to solve for $x$.