How do we solve this system of equations?

Question

$a,b \in \Bbb R$ 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?

Answer 1

$$ a=2, b=1 $$ 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.