Solving equation for integers (diophantine equations)?

Question

solve the equation for positive integers x and y

y^3=x^3+8x^2-6x+8

Answer 1

(x+3)^3

=x^3+9x^2+27x+27

>x^3+8x^2-6x+8

>x^3

so y=x+1 or y=x+2.

y=x+1

==>

x^3+3x^2+3x+1

=x^3+8x^2-6x+8,

==>

5x^2-9x+7=0, x is not a real number.

y=x+2

==>

x^3+6x^2+12x+8

=x^3+8x^2-6x+8

==>

2x^2=18x,

x=0 or x=9

so the solutions are

x=0,y=2   or

x=9,y=11

Answer 2

for two unknown you should have at least two equation

Answer 3

y^3 = x^3 + 8x^2 - 6x + 8

8x^2 - 6x + 8 = y^3 - x^3

2(4x^2 - 3x + 4) = (y - x)(y^2 + xy + x^2)

let y - x = 2, y = x + 2

4x^2 - 3x + 4 = (x+2)^2 + x(x+2) + x^2

= 3x^2 + 6x + 4

x^2 - 9x = 0

x(x - 9) = 0

x = 0 or x = 9

(x,y) = (0,2) or (9,11)

for want of positive integers x and y,

answer : x = 9, y = 11.

Answer 4

y^3=x^3+8x^2-6x+8,

y^3-x^3=8x^2-6x+8

(y-x)(y^2+xy+x^2)=2(4x^2-3x+4)

Since the discriminant of the quadratic polynomial on the right is negative, i.e. D=9-64<0 then the trinomial doesn't factor. Hence

either

y-x=2

y^2+xy+x^2=4x^2-3x+4

or

y-x=4x^2-3x+4

y^2+xy+x^2=2

The first system gives

y=x+2

y^2+xy-3x^2+3x-4=0

(x+2)^2+x(x+2)-3x^2+3x-4=0

-x^2+9x=0

x(-x+9)=0 =>x=0 or x=9 and only x=9 works for us which gives us y=11.

Consider the 2nd system.

y-x=4x^2-3x+4

y^2+xy+x^2=2

The 2-nd equation is only possible when either x or y=0 but neither is allowed by the condition that both x and y>0.

Hence the only solution in positive integeres is x=9 and y=11.