Q13/109) Find all positive integer solutions of the equation 4x^3 – 12x^2 + 5x – 10y + 36y^2 – 18y^3 + 4x^2y + 6xy – 15xy^2 = 0

We put the number in decreasing order of power of x

4x^3−12x^2 +4x^2y +5x+6xy−15xy^2−10y+36y^2−18y^3
Now factor the part independent of x
= 4x^3−12x^2 +4x^2y +5x+6xy−15xy^2−2y(5-18y+9y^2)
= 4x^3−12x^2 +4x^2y +5x+6xy−15xy^2−2y(3y-5)(3y-1)
Now I multiply by 2 and put 2x = z to get coefficient of x^3 as 1
= ½(8x^3−24x^2 +8x^2y +10x+12xy−30xy^2−4y(3y-5)(3y-1)
= 1/2(z^3−6z^2 +2 z^2y +5z+6zy−15zy^2−4y(3y-5)(3y-1)
= 1/2(z^3−z^2(6-2y) +z(5+6y+15y^2) −4y(3y-5)(3y-1)

Now we need to split - 4y(3y-5)(3y-1)into3 parts that the sum is 6- 2y

They are-( 3y-5),- (3y-1) , 4y ( it is easy to do so as 5 + 1 = 6

Now to check The coefficient of z is (5+6y+15y^2)

We see that – 4y(3y-5) – 4y(3y-1) + (3y-1) (3y-5) = -15y^2 + 6y + 5 which is true

So we get ½(( z- 4y) ( z+ 3y-5)(z+ 3y-1)) or (x-2y)(2x + 3y-5)(2x+3y-1)) = 0

Which gives us x = 2y or 2x + 3y = 5 or 2x + 3y = 1

X= 2y gives solution x= 2t, y = t;

2x + 3y = 5 gives x= 1, y= 1 and 2x + 3y = 1 has no solution ( as we are looking positive solution)