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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.

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  • $\begingroup$ 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 $\sin(x)$. $\endgroup$ May 28, 2021 at 16:27
  • $\begingroup$ @ SolubleFish, Thanks. $\endgroup$
    – user930927
    May 28, 2021 at 16:28

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Hint:
using the identity , $a+b+c=0 \iff a^3+b^3+c^3=3abc$.
On rearranging your eqation as $3.1.\beta .sinx=\beta^3 + sin^3x +1^3$
$\implies sinx +\beta +1=0$.
The rest you can do it.

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