What are the solutions to the equation m^3-27=(m-3) ?

The solutions to the equation m^3-27=(m-3) are m=3,m=-3/2+i(sqrt(23))/2 ,m=-3/2-i(sqrt(23))/2

Find the zeros of m^3-27=(m-3)

The zeros of m^3-27=(m-3) are m=3,m=-3/2+i(sqrt(23))/2 ,m=-3/2-i(sqrt(23))/2

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m^{3} - 27 = (m - 3)

m = 3, m = - \frac{3}{2} + i \frac{23}{2}, m = - \frac{3}{2} - i \frac{23}{2}

Solution steps

m^{3} - 27 = (m - 3)

Refine

m^{3} - 27 = m - 3

Move

3

to the left side

m^{3} - 24 = m

Move

m

to the left side

m^{3} - 24 - m = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + m = 0

m^{3} - m - 24 = 0

Factor

m^{3} - m - 24 : (m - 3) (m^{2} + 3 m + 8)

(m - 3) (m^{2} + 3 m + 8) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

m - 3 = 0 or m^{2} + 3 m + 8 = 0

Solve

m - 3 = 0 : m = 3

Solve

m^{2} + 3 m + 8 = 0 : m = - \frac{3}{2} + i \frac{23}{2}, m = - \frac{3}{2} - i \frac{23}{2}

The solutions are

m = 3, m = - \frac{3}{2} + i \frac{23}{2}, m = - \frac{3}{2} - i \frac{23}{2}

24 x^{3} - 24 x^{2} = 0

x^{3} - 6 x^{2} - 2 = 0

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

x^{4} - 12 x^{2} + 162 x - 12 = 0

x^{4} + 8 x^{3} + 16 x^{2} = 0

What are the solutions to the equation m^3-27=(m-3) ?
The solutions to the equation m^3-27=(m-3) are m=3,m=-3/2+i(sqrt(23))/2 ,m=-3/2-i(sqrt(23))/2
Find the zeros of m^3-27=(m-3)
The zeros of m^3-27=(m-3) are m=3,m=-3/2+i(sqrt(23))/2 ,m=-3/2-i(sqrt(23))/2