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

The solutions to the equation 3=n^3-3n^2+n are n=3,n=i,n=-i

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

The zeros of 3=n^3-3n^2+n are n=3,n=i,n=-i

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

n = 3, n = i, n = - i

Solution steps

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

Switch sides

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

Move

3

to the left side

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

Factor

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

(n - 3) (n^{2} + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

n - 3 = 0 or n^{2} + 1 = 0

Solve

n - 3 = 0 : n = 3

Solve

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

The solutions are

n = 3, n = i, n = - i

n^{2} + n^{3} = - 180

x^{3} - 27 x - 27 = 0

x^{3} + 5 x^{2} + 6 x = - 30

4 (x + 4)^{4} - 53 (x + 4)^{2} = - 49

0 = x^{3} - 2 x^{2} - 11 x + 12

What are the solutions to the equation 3=n^3-3n^2+n ?
The solutions to the equation 3=n^3-3n^2+n are n=3,n=i,n=-i
Find the zeros of 3=n^3-3n^2+n
The zeros of 3=n^3-3n^2+n are n=3,n=i,n=-i