<s>[INST] How do we solve this system of equations? $a,b \in \Bbb R$ and $$\frac{a^5b-b^5a}{a-b}=30$$ and $$a^5+b^5 =  33$$
I get that $a^6-b^6=(a-b)63$ But I have no idea how to solve after that. Someone could help me?
 [/INST] $$ a=2, b=1 $$
works. So does $$ a=1, b=2. $$
If you draw a graph of $x^5 + y^5 = C > 0$ you find that $x+y > 0.$  Thus, with $$ ab(a+b)(a^2 + b^2) = 30 $$ we find $ab > 0.$ Then, with $a^5 + b^5 = 33,$ we have both positive. 
Writing $$  a = r \cos \theta, \; b = r \sin \theta $$ and pulling out $r^5,$
we get
$$  33 \cos \theta \sin \theta (\cos \theta + \sin \theta) = 30 (\cos^5 \theta + \sin^5 \theta).   $$
Taking second derivatives, the left hand side has negative second derivative on $0 < \theta < \pi / 2.$ On the same range, the right hand side has negative first derivative until $\theta = \pi / 4,$ after which it has positive first derivative.   As a result, there are at most two points of equality, and we have already found those. 
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