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?
Please help me.

With $x=\pi$ the LHS is equal to $1$ while the RHS is equal to $-1$ therefore the identity is false.
