# How to solve the following cubic equation.

If the equation $$3\beta sinx –1 = (\beta + sinx) (\beta^2 + sin^2x – \beta sinx)$$, $$\beta \in \mathbb{R}$$ can be solved for x, then sum of all possible integral values of $$\beta$$ .

First we can use the identity $$a^3+b^3=(a+b)(a^2-ab+b^2)$$ for the left hand side of the given equation.
So it becomes $$3\beta sinx –1 =\beta^3 + sin^3x$$ Then on rearrganig we get $$sin^3x -3\beta sinx +\beta^3+1=0$$.
After this I am unable to proceed further as I how no idea how to solve the above cubic eqation or is there any onter way to do it. I though a lot about but but still could not solve. Pls help.

• Tip : when writing equation, put a backslash in front of the name of common function to get the right font and spacing. \sin(x) will give $\sin(x)$. May 28, 2021 at 16:27
• @ SolubleFish, Thanks.
– user930927
May 28, 2021 at 16:28

using the identity , $$a+b+c=0 \iff a^3+b^3+c^3=3abc$$.
On rearranging your eqation as $$3.1.\beta .sinx=\beta^3 + sin^3x +1^3$$
$$\implies sinx +\beta +1=0$$.