What are the solutions to the equation (2x+1)^3=27 ?

The solutions to the equation (2x+1)^3=27 are x=1,x=-5/4+i(3sqrt(3))/4 ,x=-5/4-i(3sqrt(3))/4

Find the zeros of (2x+1)^3=27

The zeros of (2x+1)^3=27 are x=1,x=-5/4+i(3sqrt(3))/4 ,x=-5/4-i(3sqrt(3))/4

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(2 x + 1)^{3} = 27

x = 1, x = - \frac{5}{4} + i \frac{3 3}{4}, x = - \frac{5}{4} - i \frac{3 3}{4}

Solution steps

(2 x + 1)^{3} = 27

For

g^{3} (x) = f (a)

the solutions are

Solve

The solutions are

x = 1, x = - \frac{5}{4} + i \frac{3 3}{4}, x = - \frac{5}{4} - i \frac{3 3}{4}

a, a^{3} = 64 q^{3} + 27 + 144 q^{2} + 108 q + 3

x^{4} - 8 x^{2} + 12 = 0

x^{4} + 3 x^{3} + 6 x^{2} + 3 x + 1 = 0

9 m^{2} \cdot (m - 3) = 16 (m - 3)

(8) \cdot 12 x^{4} - 56 x^{3} + 89 x^{2} - 56 x + 12 = 0

What are the solutions to the equation (2x+1)^3=27 ?
The solutions to the equation (2x+1)^3=27 are x=1,x=-5/4+i(3sqrt(3))/4 ,x=-5/4-i(3sqrt(3))/4
Find the zeros of (2x+1)^3=27
The zeros of (2x+1)^3=27 are x=1,x=-5/4+i(3sqrt(3))/4 ,x=-5/4-i(3sqrt(3))/4