Crazy solve

$y^2$ term on the RHS of the equation tells us that $y$ can not be greater than $x$ . We'll find out that, let $x-y=z$ $(y+z)^3+10(y+z)-1=y^3+6y^2 \\ (3z-6)y^2+(3z^2+10)y+z^3+10z-1=0$ Discriminant of the obtained quadratic (in terms of $y$ ) must be non-negative (why?) $(3z^2+10)^2-4(3z-6)(z^3+10z-1)\ge0 \\ 0\le z\le6$ Trying these values gives $y=5$ for $z=1$ and $y=-3$ for $z=5$ only. So totally two pairs $(x, y)=(6, 5), (2, -3)$

$\text{We may program on spread-sheet, calculator, or a computer language:-}\\Initial~~ B=C=100\\ Outer ~Do ~loop~ A=-100,1,100\\Inner~Do~Loop~x=-100,1,100\\ C=B\\\large B=\dfrac{x^3+10x-1}{A+6}-A^2 \\ If ~B=0~~ Print ~~x~and~A \\ If~C*B\leq 0~end~inner~loop~go~to~next~for~outter~loop. \\ \text{For calculator with the help of ENTRY in TI calculators, and STO} \\ \text{ things like Printing, ending loop, going to next, If, all are manual. } \\\text{We remain satisfied investigating between -100 and 100!!.}$