<s>[INST] Solving $y^3=x^3+8x^2-6x+8$ Solve for the equation $y^3=x^3+8x^2-6x+8$ for positive integers x and y.
My attempt- $$y^3=x^3+8x^2-6x+8$$
            $$\implies y^3-x^3=8x^2-6x+8$$
            $$\implies (y-x)(y^2+x^2+xy)=8x^2-6x+8$$
Now if we are able to factorise $8x^2-6x+8$ then we can compare LHS with RHS.Am I on the right track?Please help.
 [/INST] We have
$y^3 − (x + 1)^3 = x^3 + 8x^2 − 6x + 8 − (x^3 + 3x^2 + 3x + 1) = 5x^2 − 9x + 7$.
Consider the quadratic equation 
$5x^2 − 9x + 7 = 0$. 
The discriminant of this equation is
$D = 92 − 4 × 5 × 7 = −59 < 0$ and hence the expression $5x^2 − 9x + 7$ is positive for all real values of x. 
We conclude that $(x + 1)^3 < y^3$ and hence $x + 1 < y$.
On the other hand we have
$(x + 3)^3 − y^3 = x^3 + 9x^2 + 27x + 27 − (x^3 + 8x^2 − 6x + 8) = x^2 + 33x + 19 > 0$
for all positive $x$.
We conclude that $y < x + 3$. 
Thus we must have $y = x + 2$. Putting this value of y, we get $0 = y^3 − (x + 2)^3 = x^3 + 8x^2 − 6x + 8 − (x^3 + 6x^2 + 12x + 8) = 2x^2 − 18x$.
We conclude that $x = 0$ and $y = 2$ or $x = 9$ and $y = 11$
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