What are the solutions to the equation n^2(n^2+n+1)=0 ?

The solutions to the equation n^2(n^2+n+1)=0 are n=0,n=-1/2+i(sqrt(3))/2 ,n=-1/2-i(sqrt(3))/2

Find the zeros of n^2(n^2+n+1)=0

The zeros of n^2(n^2+n+1)=0 are n=0,n=-1/2+i(sqrt(3))/2 ,n=-1/2-i(sqrt(3))/2

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n^{2} (n^{2} + n + 1) = 0

n = 0, n = - \frac{1}{2} + i \frac{3}{2}, n = - \frac{1}{2} - i \frac{3}{2}

Solution steps

n^{2} (n^{2} + n + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

n = 0 or n^{2} + n + 1 = 0

Solve

n^{2} + n + 1 = 0 : n = - \frac{1}{2} + i \frac{3}{2}, n = - \frac{1}{2} - i \frac{3}{2}

The solutions are

n = 0, n = - \frac{1}{2} + i \frac{3}{2}, n = - \frac{1}{2} - i \frac{3}{2}

s^{3} + 2 s^{2} - s - 2 = 0

(3 x - 1)^{6} = 1

x^{3} (2 x^{2} + 24) = 128 x

(x - 7)^{3} = 8

x^{2} 8 x - 4 = 0

What are the solutions to the equation n^2(n^2+n+1)=0 ?
The solutions to the equation n^2(n^2+n+1)=0 are n=0,n=-1/2+i(sqrt(3))/2 ,n=-1/2-i(sqrt(3))/2
Find the zeros of n^2(n^2+n+1)=0
The zeros of n^2(n^2+n+1)=0 are n=0,n=-1/2+i(sqrt(3))/2 ,n=-1/2-i(sqrt(3))/2