how to find a (x1^3n+x2^3n) if x1 and x2 are the root of quadratic equation x^2+x+1
To find the value of (x1^3n + x2^3n) given that x1 and x2 are the roots of the quadratic equation x^2 + x + 1, we need to use Vieta's formulas.
Vieta's formulas state that for a quadratic equation in the form of ax^2 + bx + c = 0, if the roots are x1 and x2, then the following relationships hold:
x1 + x2 = -b/a
x1 * x2 = c/a
In this case, the given quadratic equation is x^2 + x + 1 = 0.
Comparing it with the general form ax^2 + bx + c = 0, we have:
a = 1, b = 1, c = 1.
According to Vieta's formulas, we know:
x1 + x2 = -b/a = -1/1 = -1
x1 * x2 = c/a = 1/1 = 1
Now, let's simplify the expression (x1^3n + x2^3n):
(x1^3n + x2^3n) = (x1^n * x1^2 + x2^n * x2^2)
= (x1^n * (x1^2) + x2^n * (x2^2))
Since x1 and x2 are the roots of the quadratic equation x^2 + x + 1 = 0, we can rewrite the equation as follows:
x1^2 + x1 + 1 = 0
x2^2 + x2 + 1 = 0
Now let's substitute (x1^2 + x1 + 1) for -x1 and (x2^2 + x2 + 1) for -x2:
(x1^3n + x2^3n) = (x1^n * (-x1) + x2^n * (-x2))
= (-x1^(n+1) - x2^(n+1))
So, the value of (x1^3n + x2^3n) is -x1^(n+1) - x2^(n+1), where x1 and x2 are the roots of the quadratic equation x^2 + x + 1.