# What is x^3+y^3+z^3, if x,y,z are the solutions of the equation x^3-2x^2+2x+17=0?

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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If x,y,z are the roots of the given equation, then substituted within equation, they verify it.

`x^3-2x^2+2x+17=0`

`y^3-2y^2+2y+17=0`

`z^3-2z^2+2z+17=0`

`x^3-2x^2+2x+17+y^3-2y^2+2y+17+z^3-2z^2+2z+17=0`

We'll isolate to the left side the sum of cubes:

`x^3 + y^3 + z^3 = 2x^2 - 2x - 17 + 2y^2 - 2y - 17 + 2z^2 - 2z - 17`

`x^3 + y^3 + z^3 = 2(x^2 + y^2 + z^2) - 2(x + y + z) - 3*17`

We'll use Viete's relations to determine the sum of the roots and the sum of the squares of the roots.

`x + y + z = -(-2)/1 = 2`

`x^2 + y^2 + z^2 = (x + y + z)^2 - 2(xy + xz + yz)`

We'll use Viete's relations again to calculate the sum of the products of two roots:

`xy + xz + yz = 2/1 = 2`

`` `x^2 + y^2 + z^2 = (2)^2 - 2*2 = 4 - 4 = 0`

`x^3 + y^3 + z^3 = 2*(0) - 2*(2) - 3*17`

`x^3 + y^3 + z^3 = -4 - 51`

`` `x^3 + y^3 + z^3 = -55`

Therefore, the sum of the cubes of the roots of equation is:

`x^3 + y^3 + z^3 = -55`