# What is the sum of the cubes of the roots of equation x^2=7-5x?

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By definition, the roots of an equation, substituted in the equation, cancel out the equation.

We'll move all terms to one side:

x^2 + 5x - 7 = 0

We'll substitute x1 and x2 in the original equation:

x1^2 + 5x1 - 7 = 0 (1)

x2^2 + 5x2 - 7 = 0 (2)

We'll add (1) and (2):

(x1^2 + x2^2) + 5(x1 + x2) - 14 = 0

We'll use Viete's relations to express the sum x1 + x2:

x1 + x2 = -5

(x1^2 + x2^2) + 5*(-5) - 14 = 0

(x1^2 + x2^2) - 25 - 14 = 0

We'll combine like roots:

(x1^2 + x2^2) - 39 = 0

We'll add 39 both sides:

(x1^2 + x2^2) = 39

Now, we'll re-write the equations (1) and (2):

x1^2 + 5x1 - 7 = 0 (1)

x2^2 + 5x2 - 7 = 0 (2)

We'll multiply (1) by x1 and (2) by x2:

x1^3 + 5x1^2 - 7x1 = 0 (3)

x2^3 + 5x2^2 - 7x2 = 0 (4)

We'll add (3) and (4):

(x1^3 + x2^3) + 5(x1^2 + x2^2) - 7(x1 + x2) = 0

(x1^3 + x2^3) = 7(x1 + x2) - 5(x1^2 + x2^2)

(x1^3 + x2^3) = 7*(-5) - 5*(39)

(x1^3 + x2^3) = -35 - 195

(x1^3 + x2^3) = 230

**The sum of the cubes of the roots of the given equation is(x1^3 + x2^3) = 230.**