# x1 and x2 are solutions of x^2-x+1=0. What is the value of the sum (x1^2-x1)^1000 + (x2^2-x2)^1000 ?

*print*Print*list*Cite

### 2 Answers

x1 and x2 are given as the solutions of x^2 - x + 1 = 0

So if we substitute x with x1 we get x1^2 - x1 + 1 = 0

=> x1^2 - x1 = -1

Similarly with x2, x2^2 - x2 + 1 = 0

=> x2^2 - x2 = -1

The required sum is (x1^2-x1)^1000 + (x2^2-x2)^1000

=> -1^1000 + -1^1000

=> 1 + 1

=> 2

**The required sum is 2.**

We know the fact that the solution of an equation, substituted into equation, it cancels out the equation.

We'll substitute the solutions of the equation into the given equation:

x1^2 - x1 + 1 = 0

x1^2 - x1 = -1We'll raise to the power of 1000 both sides:

(x1^2-x1)^1000 = (-1)^1000

(x1^2-x1)^1000 = 1 (1)

x2^2 - x2 + 1 = 0

x2^2 - x2 = -1We'll raise to the power of 1000 both sides:

(x2^2-x2)^1000 = (-1)^1000

(x2^2-x2)^1000 = 1 (2)

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

(x1^2-x1)^1000 + (x2^2-x2)^1000 = 1 + 1

(x1^2-x1)^1000 + (x2^2-x2)^1000 = 2

**The value of the sum = 2.**