<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] We will use a few identities.
First is the Pythagorean Identity:
$$\sin^2\alpha +\cos^2\alpha = 1$$
Second is a simple factorization:
$$a^5 - b^5 = (a-b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4) = (a-b)(a^4 + b^4 + ab(a^2+b^2+ab))$$
Third will help us write things of fourth degree in terms of second and first degree terms:
$$a^4 + b^4 = (a^2+b^2)^2-2(ab)^2$$
So we can combine them and get the following:
$$a^5 - b^5 = (a-b)((a^2+b^2)^2-2(ab)^2 + ab(a^2+ab+b^2))$$
Now, all we have to do is calculate $\sin\alpha \cos\alpha$ and plug the values in.
$$\sin\alpha - \cos\alpha = \frac{1}{2}$$
Squaring both sides, we get
$$\sin^2\alpha + \cos^2\alpha - 2\sin\alpha\cos\alpha = \frac{1}{4}$$
$$\Rightarrow 1 - 2\sin\alpha\cos\alpha = \frac{1}{4}$$
$$\Rightarrow\sin\alpha\cos\alpha = \frac{3}{8}$$
Then, if we let $a = \sin\alpha$ and $b = \cos\alpha$, we get
$$\sin^5\alpha - \cos^5\alpha = (\sin\alpha - \cos\alpha)((\sin^2\alpha + \cos^2\alpha)^2 - 2(\sin\alpha\cos\alpha)^2 \\+ \sin\alpha\cos\alpha (\sin^2\alpha + \cos^2\alpha + \sin\alpha\cos\alpha))$$
$$= \frac{1}{2}\left(1^2 - 2\left(\frac{3}{8}\right)^2 + \frac{3}{8}\left(1+\frac{3}{8}\right)\right) = \frac{79}{128}$$
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