You should notice that the bottom equation is a summation of cubes, hence, you may use the following special product, such that:

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

Replacing `2` for `x + y` yields:

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

Since...

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You should notice that the bottom equation is a summation of cubes, hence, you may use the following special product, such that:

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

Replacing `2` for `x + y` yields:

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

Since `x^3 + y^3 = 2` yields:

`2 = 2(x^2 - xy + y^2)`

Reducing duplicate factors both sides, yields:

`1 = x^2 - xy + y^2`

You need to use the following identity, such that:

`x^2 + y^2 = (x + y)^2 - 2xy`

Replacing `2` for `x + y` , yields:

`x^2 + y^2 = 2^2 - 2xy`

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

Replacing `4 - 2xy` for `x^2 + y^2` in equation `1 = x^2 - xy + y^2` , yields:

`1 =4 - 2xy - xy => 1 - 4 = -3xy => -3 = -3xy => xy = 1`

You need to use Lagrange resolvents to evaluate `x` and `y` , such that:

`z^2 - (x + y)*z + xy = 0`

Replacing 2 for `x + y` and 1 for `xy` yields:

`z^2 - 2z + 1 = 0 => (z - 1)^2 = 0 => z_1 = z_2 = 1`

**Hence, evaluating `x` and `y` , under the given conditions, yields **`x = y = 1.`

**Further Reading**