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 ?
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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.
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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.
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