# If x1 and x2 are the roots of equation x^2+x+1=0, find x1^10+x2^10=?

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### 2 Answers

x^2+x+1 = 0.

To find the x1^10 +x2^10

By the relation between the roots and coefficients, x1+x2 = -1/1 = -1.

where x1 and x2 are the roots of the given equation

Multiply the equation by x-1.

Then (x-1)(x^2+x+1) = 0

x^3-1 = 0

So x^2+x+1 = 0 has the cube roots of unity w = (-1+sqrt(-3))/2 and w^2 = (-1-sqrt(-3))/2.

Therefore w ^3 = 1.

Therefore

x1^10 +x2^10 = w^10 + (w^2)^10 = (w^9)*w +(w^18)*w^2

= (w^3)^3 * w + (w^3)^6 * w^2

= 1^3*w +1^6 *w^2

= x1+x2

= - 1/1

= -1

Therefore x1^10+x2^10 = -1

If x^2 + x + 1 = 0, that means that if we'll multiply both sides by (x-1), we'll get:

(x-1)(x^2 + x + 1) = 0

But the product is the result of difference of cubes:

x^3 - 1 = (x-1)(x^2 + x + 1)

If (x-1)(x^2 + x + 1) = 0, then x^3 - 1 = 0

We'll add 1 both sides:

x^3 = 1

x1 and x2 are the roots of the equation x^3 - 1 = 0, so:

x1^3 = 1

x2^3 = 1

We'll write x1^10 = x1^(9+1)

x1^(9+1) = x1^9*x1

x1^9*x1 = (x1^3)^2*x1

But x1^3 = 1

x1^10 = 1^2*x1

x1^10 = x1

x2^10 = x2

So, x1^10 + x2^10 = x1 + x2

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

x1 + x2 = -b/a

where b and a are the coefficients of the quadratic equation

ax^2 + bx + c = 0

In our case, the quadratic is:

x^2 +x +1 = 0

a = 1

b = 1

x1 + x2 = -1/1

x1 + x2 = -1

**x1^10 + x2^10 = -1**