<s>[INST] Calculate $\sin^5\alpha-\cos^5\alpha$ if $\sin\alpha-\cos\alpha=\frac12$ Calculate $$\sin^5\alpha-\cos^5\alpha$$ if $\sin\alpha-\cos\alpha=\dfrac12$.
The main idea in problems like this is to write the expression that we need to calculate in terms of the given one (in this case we know $\sin\alpha-\cos\alpha=\frac12$).
I don't see how to even start to work on the given expression as we cannot use $a^2-b^2=(a-b)(a+b)$ or $a^3-b^3=(a-b)(a^2+ab+b^2)$. So in other words, I can't figure out how to factor the expression (even a little).

 The given answer is $\dfrac{79}{128}$.

 [/INST] $$\dfrac1{2\sqrt2}=\sin\left(\alpha-\dfrac\pi4\right)=\sin x$$ where $\alpha-\dfrac\pi4=x$
$$\sin^5\alpha-\cos^5\alpha=\sin^5\left(x+\dfrac\pi4\right)-\cos^5\left(x+\dfrac\pi4\right)=\dfrac{(\cos x+\sin x)^5-(\cos x-\sin x)^5}{(\sqrt2)^5}$$
Now,
$$(\cos x+\sin x)^5-(\cos x-\sin x)^5=2\sin^5x+2\binom53\sin^3x\cos^2x+2\binom54\sin x\cos^4x$$
Use $\cos^2x=1-\sin^2x=1-\left(\dfrac1{2\sqrt2}\right)^2=?$
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