What are the solutions to the equation (x^2+2x-14)^5=1 ?

The solutions to the equation (x^2+2x-14)^5=1 are x=3,x=-5

Find the zeros of (x^2+2x-14)^5=1

The zeros of (x^2+2x-14)^5=1 are x=3,x=-5

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(x^{2} + 2 x - 14)^{5} = 1

x = 3, x = - 5

Solution steps

(x^{2} + 2 x - 14)^{5} = 1

For

(g (x))^{n} = f (a)

, n is odd, the solution is

Solve

The solutions are

x = 3, x = - 5

m^{3} - 2 m + 1 = 0

\frac{z ^{3}}{2} = 8 i

x, a = (- \frac{1}{2} \cdot x^{2} \cdot y^{3})^{1} \cdot (- x \cdot y^{2})

2 x^{3} - 2 x^{2} + 2 x - 5 = 0

x^{3} - 4 x^{2} - x + 1 = (x - 2) 3

What are the solutions to the equation (x^2+2x-14)^5=1 ?
The solutions to the equation (x^2+2x-14)^5=1 are x=3,x=-5
Find the zeros of (x^2+2x-14)^5=1
The zeros of (x^2+2x-14)^5=1 are x=3,x=-5