$a= \frac{b^2-2b+16}{b^3} b= \frac{8a^3}{a^2-2a+16} a=? b=?$

Question

it's part of my bigger homework, i got to state like that and i need some nice way to calculate a and b. I was trying to do this by setting b value from second equations to first equations, but it feels like so much work, can i do it smarter way? If it's impossible i will send all my exercise, so you can see if i didn't noticed a better way to do whole thing. $$ a= \frac{b^2-2b+16}{b^3}$$ $$ b= \frac{8a^3}{a^2-2a+16}$$

Answer 1

HINT:

$8 a^3 b^3= 8 a^2(b^2-2b+16) = b^4(a^2-2a+16)$.

setting $a = x b$ you have

$8 x^3 b^6= 8 x^2 b^2(b^2-2b+16) = b^4(x^2 b^2-2 x b +16)$

Equating the second and third expression gives a quadratic equation in $x$ which you can solve for $x = x (b)$.

Equating the first and second expression gives $ x = \frac{b^2-2b+16}{b^4}$ which you can now equate with your $x (b)$ obtained above, to obtain $b$. Here the story may become nasty....