What are the solutions to the equation n^2=n^3+n^1 ?

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

Find the zeros of n^2=n^3+n^1

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

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n^{2} = n^{3} + n^{1}

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

Solution steps

n^{2} = n^{3} + n^{1}

Switch sides

n^{3} + n^{1} = n^{2}

Move

n^{2}

to the left side

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

Factor

n^{3} + n - n^{2} : n (n^{2} - n + 1)

n (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}

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

23305 = 7000 (1 + x)^{3}

(6 x^{2} - x - 2) (a x + 9) = 0, a \neq = 0

16 x^{3} + 64 x^{2} - 25 x - 100 = 0

2 x^{4} - 4 x^{2} + 1 = 0

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