If x,y,z are the roots of the given equation, then substituted within equation, they verify it.
We'll add the equations above:
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`