<s>[INST] How can I prove that $2(\cos^6(x)-\sin^6(x))-3(\cos^4(x)+\sin^4(x))=-4\sin^6(x)-1$ How can I prove that $2(\cos^6(x)-\sin^6(x))-3(\cos^4(x)+\sin^4(x))=-4\sin^6(x)-1$
I tried to factor and I got $2\cos^4(x)+(-2\sin^2(x)-3)(\cos^4(x)+\sin^4(x))$ but that doesn't lead me to my goal.
I also tried to write all the cosines in terms of sines  to have: $-3\sin^6(x)-9\sin^2(x)-2-3\sin^4(x)$
But I don't see how to continue
Any hint is welcome! thnxx 
 [/INST] Apply $a^2+b^2=(a+b)^2-2ab$ in $$\sin^4x+\cos^4x=(\sin^2x)^2+(\cos^2x)^2$$
and $a^3+b^3=(a+b)^3-3ab(a+b)$ in $$\sin^6x+\cos^6x=(\sin^2x)^3+(\cos^2x)^3$$
Finally the re-arrange the terms
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