# Prove that x1^3+x2^3 is an integer.If x1 and x2 are the solutions of x^2+5x-7=0.

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You should also use the sum of two cubes formula, such that:

`a^3 + b^3 = (a + b)(a^2 - ab + b^2)`

Reasoning by analogy yields:

`x_1^3 + x_2^3 = (x_1 + x_2)(x_1^2 - x_1*x_2 + x_2^2)`

You may find the solutions `x_1,x_2` to quadratic equation, using the quadratic formula, such that:

`x_(1,2) = (-5+-sqrt(25 + 28))/2`

`x_(1,2) = (-5+-sqrt53)/2`

You need to evaluate `x_1 + x_` 2 such that:

`x_1 + x_2 = (-5 + sqrt53 - 5 - sqrt53)/2`

`x_1 + x_2 = -5`

You need to evaluate `x_1*x_2` such that:

`x_1*x_2 = ((-5 + sqrt53)(-5 - sqrt53))/4`

`x_1*x_2 = (25 - 53)/4 => x_1*x_2 = -7`

`x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1*x_2`

`x_1^2 + x_2^2 = 25 + 14 => x_1^2 + x_2^2 = 39`

Substituting -5 for `x_1 + x_2 , -7` for x`_1*x_2` and 39 for `x_1^2 + x_2^2` , yields:

`x_1^3 + x_2^3 = (-5)(39 + 7) => x_1^3 + x_2^3 =-230`

**Hence, evaluating the sum of cubes yields `x_1^3 + x_2^3 = -230` .**