Prove $\sin^3A-\cos^3A=\left(\sin^2A-\cos^2A\right)(1-2\sin^2A\cos^2A)$

Prove $$\sin^3A-\cos^3A=\left(\sin^2A-\cos^2A\right)(1-2\sin^2A\cos^2A)$$

My attempt is as follows:

Taking LHS:

$$\left(\sin A-\cos A\right)(1+\sin A\cos A)$$ $$\left(\sin^2A-\cos^2A\right)\frac{\left(1+\sin A\cos A\right)}{\left(\sin A+\cos A\right)}$$ $$\left(\sin^2A-\cos^2A\right)\frac{(\left(\sin A+\cos A\right)^2-\sin A\cos A)}{\sin A+\cos A}$$ $$\left(\sin^2A-\cos^2A\right)\left(\sin A+\cos A-\frac{\sin A\cos A}{\sin A+\cos A}\right)$$

I was not getting any breakthroughs from here.

So I tried RHS:

$$(\sin A-\cos A)(\sin A+\cos A)(1-2\sin^2A\cos^2A)$$ $$(\sin A-\cos A)(\sin A+\cos A)((\sin^2A+\cos^2A)^2-2\sin^2A\cos^2A)$$ $$(\sin A-\cos A)(\sin A+\cos A)(\sin^4A+\cos^4A)$$

Even from here I was not getting breakthroughs, what am i missing?

• Hint:$$\sin^3A-\cos^3A=\sin^3A-\sin^2A\cos A+\sin^2A\cos A-\sin A\cos^2A+\sin A\cos^2A-\cos^3A$$ $$=\sin^2A(1-\cos A)+\sin A\cos A(\sin A-\cos A)-\cos^2A(1-\sin A)$$ – Don Thousand Oct 20 at 16:21
With $$x=\pi$$ the LHS is equal to $$1$$ while the RHS is equal to $$-1$$ therefore the identity is false.
$$(\sin^2A-\cos^2A)(1-\sin^2A\cos^2A)$$ $$=(\sin A-\cos A)(\sin A+\cos A)(\sin^4A+\cos^4A).$$
But I think the problem is wrong, just try with $$A=\frac{\pi}{5}$$.