…. *y^3 = x^3 + 8x^2 – 6x + 8* for positive integers x and y.

There are many ways to approach this problem. One of the ways is to assume

*y = x + a*, for all *a* belongs to *z*

The problem can then be reduced to a quadratic expression in x. The discriminant then contains a fourth order expression in a. It gets tricky from there. So, a better approach to solve it is through reduction of available range of x and y. To start with, x and y could be in the range 0 to infinity.

*y^3 = x^3 + 8x^2 – 6x + 8 ….. (1)*

*y^3 = (x^3 + 6x^2 + 12x + 8) + (2x^2 – 18x)*

*y^3 = (x^3 + 6x^2 + 12x + 8) + 2x(x – 9) *

*y^3 = (x+2)^3 + 2x(x – 9)….. (2)*

*y > 0 => y^3 >0*

The equation reduces to a perfect cube when

*2x(x-9) = 0 ….. (3)*

*x=0 or x=9*

We have to now find if there could be any others values in range that can satisfy (1). To do this,

assume *x = y + a where a is an integer*

From (2),

*y^3 – 2x(x – 9) = (x + 2)^3 *

*y^3 – 2y^2 + y(-4a + 18) + (18a – 2a^2) = (x+2)^3 …. (5)*

If LHS of (5) were to be a perfect cube, then

*((coefficient of y^2)/3) x ((coefficient of y)/3) = (constant)^2*

This reduces to the quadratic equation

*9a^2 – 77a – 18 = 0 *

Neither of the roots of this quadratic equation is a positive integer.

It follows from this and (4) that only possible values of x are 0 and 9. For x = 0, y = 2 and for x = 9, y = 11. Since x and y are positive integers, in strict sense, x = 9 and y = 11 is only possible solution such that x and y are positive integers.

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