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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.

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  • $\begingroup$ Hint:$$\sin^3A-\cos^3A=\sin^3A-\sin^2A\cos A+\sin^2A\cos A-\sin A\cos^2A+\sin A\cos^2A-\cos^3A$$ $$=\sin^2A(1-\cos A)+\sin A\cos A(\sin A-\cos A)-\cos^2A(1-\sin A)$$ $\endgroup$ – Don Thousand Oct 20 at 16:21
  • $\begingroup$ this expression is also not leading anywhere, I tried it $\endgroup$ – user3290550 Oct 20 at 16:28
  • $\begingroup$ please see what I have done in my attempt in the question $\endgroup$ – user3290550 Oct 20 at 16:29
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With $x=\pi$ the LHS is equal to $1$ while the RHS is equal to $-1$ therefore the identity is false.

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

But I think the problem is wrong, just try with $A=\frac{\pi}{5}$.

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