Solving equation $x^3+x^2+x=1$

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Find a solution for equation $x^3+x^2+x=1$.

I found $x=0.54369 (which is approximately equal to Cosine of 57 degree) by

trial and error as follows:

$(0.6)^3+(0.6)^2+0.6=1.176$...

$(0.5)^3+(0.5)^2+0.5=0.875$...

so $0.5< X < 0.6$

and $(0.54369)^3+(0.54369)^2+ 0.54369=1.000002936$

can anyone gives an anlytic solution for this equation?

Thanks for solution, it could be more useful if it was in detail.

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If you use Cardano's method, you will notice that the equation has only a real root.

Using the depressed equation $$t^3+pt+q=0$$ you have $p=\frac{2}{3}$ and $q=-\frac{34}{27}$. Looking at the hyperbolic solution for one real root leads to $$t=\frac{2\sqrt{2}}{3} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{17}{2 \sqrt{2}}\right)\right)$$ Back to $x$ the solution write $$x=-\frac 13+\frac{2\sqrt{2}}{3} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{17}{2 \sqrt{2}}\right)\right)$$ which is $\approx 0.5436890$ while $\cos \left(\frac{57\pi }{180}\right)\approx 0.54463904$