Solve Sin(^6)x + Cos(^6)x Factoring Problem

[riddlingminion]

How do I factor sin(^6)x + cos(^6)x?

 

[lotrgreengrapes7926]

Use x^3+y^3=(x+y)(x^2-xy-y^2).

 

[dextercioby]

Better

$$\sin^{6}x +\cos^{6}x= (\sin^{2}x +\cos^{2}x)(\sin^{4}x +\cos^{4}x)-2\sin^{2}x \cos^{2}x(\sin^{2}x +\cos^{2}x)=...=\cos^{2} 2x.$$

Daniel.

 

[lotrgreengrapes7926]

Doesn't $$( \sin^{2} x +\cos^{2} x ) ( \sin^{4}x +\cos^{4} x ) - 2 \sin^{2}x \cos^{2} x ( \sin^{2} x +\cos^{2} x) = \sin^6 x+ \cos^6 x - \sin^2 x \cos^4 x- \sin^4 x \cos^2 x$$?

 

[arildno]

Better

$$\sin^{6}x +\cos^{6}x= (\sin^{2}x +\cos^{2}x)(\sin^{4}x +\cos^{4}x)-2\sin^{2}x \cos^{2}x(\sin^{2}x +\cos^{2}x)=...=\cos^{2} 2x.$$

Daniel.

Whereby the hitherto unknown equality:
$$\frac{1}{8}+\frac{1}{8}=0$$
is proven.

 

[dextercioby]

Yes, thought it was 2 simple 2 be true. You can drop that 2, then. :d

Daniel.