<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] Alternative approach:
Let $~\displaystyle x = \sin(\alpha), y = \cos(\alpha) = \implies 
(x - y) = \frac{1}{2}, ~\left(x^5 - y^5\right) = \color{red}{what?}$
Tools

*

*$T_1 ~: ~(x - y)^2 = x^2 - 2xy + y^2.$


*$T_2 ~: ~(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3.$


*$T_3 ~: ~(x - y)^5 = x^5 - 5x^4y + 10x^3y^2 - 10x^2y^3 + 5xy^4 - y^5.$
Plan of Attack 
Since $~x^2 + y^2 = 1~,$ there is sufficient information to compute $(xy)$.  Using this you can compute $x^3 - y^3$, and using this you can compute $x^5 - y^5.$

From $T_1,$ you have that
$$\frac{1}{4} = 1 - 2xy \implies xy = \frac{3}{8}. \tag1 $$

Using (1), from $T_2,$ you have that
$$\frac{1}{8} = \left(x^3 - y^3\right) - 3xy(x-y) \implies $$
$$\frac{1}{8} = \left(x^3 - y^3\right) - 3\left(\frac{3}{8}\right) \left(\frac{1}{2}\right) \implies (x^3 - y^3) = \frac{11}{16}. \tag2 $$

Using (1) and (2), from $T_3,$ you have that
$$\frac{1}{32} = \left(x^5 - y^5\right) -5(xy)(x^3 - y^3) + 10(xy)^2(x - y) \implies $$
$$\frac{1}{32} = \left(x^5 - y^5\right) - 5\left(\frac{3}{8}\right)\left(\frac{11}{16}\right) + 10\left(\frac{3}{8}\right)^2\left(\frac{1}{2}\right) \implies $$
$$\frac{4}{128} = \left(x^5 - y^5\right) - \frac{165}{128} + \frac{90}{128} \implies$$
$$\left(x^5 - y^5\right) = \frac{4 + 165 - 90}{128} = \frac{79}{128}.$$
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