What are the solutions to the equation (5-2x)^3=-27 ?

The solutions to the equation (5-2x)^3=-27 are x=4,x= 7/4+i(3sqrt(3))/4 ,x= 7/4-i(3sqrt(3))/4

Find the zeros of (5-2x)^3=-27

The zeros of (5-2x)^3=-27 are x=4,x= 7/4+i(3sqrt(3))/4 ,x= 7/4-i(3sqrt(3))/4

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(5 - 2 x)^{3} = - 27

x = 4, x = \frac{7}{4} + i \frac{3 3}{4}, x = \frac{7}{4} - i \frac{3 3}{4}

Solution steps

(5 - 2 x)^{3} = - 27

Expand

(5 - 2 x)^{3} : 125 - 150 x + 60 x^{2} - 8 x^{3}

125 - 150 x + 60 x^{2} - 8 x^{3} = - 27

Move

27

to the left side

- 8 x^{3} + 60 x^{2} - 150 x + 152 = 0

Factor

- 8 x^{3} + 60 x^{2} - 150 x + 152 : - 2 (x - 4) (4 x^{2} - 14 x + 19)

- 2 (x - 4) (4 x^{2} - 14 x + 19) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 4 = 0 or 4 x^{2} - 14 x + 19 = 0

Solve

x - 4 = 0 : x = 4

Solve

4 x^{2} - 14 x + 19 = 0 : x = \frac{7}{4} + i \frac{3 3}{4}, x = \frac{7}{4} - i \frac{3 3}{4}

The solutions are

x = 4, x = \frac{7}{4} + i \frac{3 3}{4}, x = \frac{7}{4} - i \frac{3 3}{4}

3 x^{4} = 9

x^{4} - 9 = 2 x^{4} + 6 x^{2}

x^{3} - 14 x^{2} - 49 = 0

2 x^{3} - 6 x + 5 = 0

(1.06)^{40} \cdot 0.5 = x^{40}

What are the solutions to the equation (5-2x)^3=-27 ?
The solutions to the equation (5-2x)^3=-27 are x=4,x= 7/4+i(3sqrt(3))/4 ,x= 7/4-i(3sqrt(3))/4
Find the zeros of (5-2x)^3=-27
The zeros of (5-2x)^3=-27 are x=4,x= 7/4+i(3sqrt(3))/4 ,x= 7/4-i(3sqrt(3))/4