<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] using the hint: $x^5-y^5 = (x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)$
Let's create the parts:
since $x^2+y^2=1$ we have:
$(x^2+y^2)^2 = x^4+2x^2y^2+y^4 = 1$ (2)
We have a big part of it but we are missing $xy$ so
we know $x-y = 1/2$ squaring both side: $x^2 + y^2 -2xy = 1/4$ since $x^2+y^2=1$ then $xy=3/8$ (1).
using (2)
$(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4) = \frac{1}{2} (1 - x^2y^2 + xy(x^2+y^2)) = \frac{1}{2} (1 - \frac{9}{64} + \frac{3}{8}\times1) = \frac{79}{128}$
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