You should use the following special product in top equation, such that:

`(a - b)(a^2 + ab + b^2) = a - b`

Reasoning by analogy, you should multiply the top equation, both sides, by the factor `(root(3)(x^2) + root(3)(xy) + root(3)(y^2))` , such that:

`(root(3)x - root(3)y)(root(3)(x^2) + root(3)(xy) + root(3)(y^2)) = 0 => x - y = 0`

Replacing `x - y = 0` for `(root(3)x - root(3)y) = 0` in top equation yields:

`{(x - y = 0),(x + y = 2):}`

You need to add the equations to remove the variable y, such that:

`2x = 2 => x = 1`

Replacing 1 for x in the bottom equation yields:

`1 + y = 2 => y = 2 - 1 => y = 1`

**Hence, evaluating the solution to simultaneous equations yields **`x = 1, y = 1.`

`root3x-root3y=0`

`root3x=root 3 y`

Thus

raise power three both side

`(root 3x)^3=(root 3y)^3`

`x=y` (i)

also

`x+y=2` (ii)

Thus from (i),substitute value of y in (ii)

`x+x=2`

`2x=2`

`x=1`

`so y=1`

**Thus answer is**

**x=y=1**