Solving a cubic equation involving trigonometric functions Question:
If $$\sin(x) - \cos(x) = \frac{\sqrt{3}}2$$
Then $$\sin^3(x) - \cos^3(x) = ?$$
I have turned first equation into a quadratic so I got $$\sin(x) = \frac{\sqrt{3}\mp\sqrt{5}}4$$ and $$\cos(x) = \frac{-\sqrt{3}\mp\sqrt{5}}4$$ But stuck here. I don't know what should I do, please help

$$\sin(x) - \cos(x) = \frac{\sqrt{3}}2$$
$$\implies \sin^2(x) + \cos^2(x)-2\sin x \cos x = \dfrac 34$$
$$\implies \sin x\cos x=\dfrac 18$$ (using $\sin^2x+\cos^2x=1$)
Now, using $a^3-b^3=(a-b)(a^2+ab+b^2)$,$$\sin^3(x)-\cos^3x = (\sin x-\cos x)(\sin^2x+\cos^2x+\sin x\cos x)$$
$$=\dfrac {\sqrt 3}{2}\times \left (1+\dfrac 18\right)$$
$$=\dfrac {9\sqrt 3}{16}$$
