Find the ordered pair (x,y) such that
$x + y ^3 = 100$
$x^2 - y^2 = 100$
I decided to substitute $x = 100 - y^3$ into the second equation and got
$10000 - 200y^3 + y^6 - y^2 = 100$
Is there any trick or method I can use to solve the 6th degree polynomial above? Any insights would be appreciated.
Beside numerical methods, you could "easily" estimate the solutions.
Consider that you look for the zero's of function $$f(y)=y^6-200 y^3-y^2+9900$$ for which $$f'(y)=2 y \left(3 y^4-300 y-1\right)$$ If you neglect the $-1$, you have a solution which is "close" to $y=10^{2/3}$.
Make a series expansion of $f(y)$ around this value $$f(y)=-10 \left(10+10^{1/3}\right)-2\ 10^{2/3} \left(y-10^{2/3}\right)+\left(900\ 10^{2/3}-1\right) \left(y-10^{2/3}\right)^2+O\left(\left(y-10^{2/3}\right)^3\right)$$ which is a quadratic in $\left(y-10^{2/3}\right)$. Solve it to get $$y_\pm=10^{2/3}+\frac{10^{2/3}\pm 10 \sqrt{899+900\ 10^{2/3}}}{900\ 10^{2/3}-1}$$ which gives $y_-\sim 4.4721$ and $y_+\sim 4.8133$ while the exact solution would be $4.4656$ and $4.8073$.
