Question

Solve the equation
z³=27

Answer 1

Answer: z = 3, -1.5 + 1.5
`√(3)`i, -1.5 - 1.5
`√(3)`i

Explanation:

A cubic equation will have three solutions. This equatoin will give us one real solution. We will also have two complex solutions. These are solutions that involve "imaginary numbers" and will utilize the value i. We will write the equation in standard form and factor it.

z³ - 27 = 0

(z - 3)(z² + 3z + 9) = 0

The first factor gives us z = 3, because 3 - 3 = 0. Now, we will look at when the second factor equals 0. Let us complete the square.

z² + 3z + 9 = 0

(z² + 3z + 2.25) + 9 - 1.5 = 0

(z² + 3z + 2.25) + 6.75 = 0

(z + 1.5)² + 6.75 = 0

Lastly, we can continue solving. We will subtract 7.5 from both sides of the equation and then take the square root of both sides. Then, we will subtract 1.5 from both sides of the equation and simplify the radical.
`i=√(-1)`.

(z + 1.5)² = -6.75

z + 1.5 = ±
`√(6.75)`i

z = -1.5 ±1.5
`√(3)`i

Answer 2

z = 3, -1.5+1.5√3i, -1.5-1.5√3i

Explanation:

You want the solutions to the equation z³ = 27.

Cubic

A cubic equation has three solutions. This one has one positive real solution and two complex conjugate solutions.

We can rewrite the equation to standard form, then factor the difference of cubes.

z³ -27 = 0

(z -3)(z² +3z +9) = 0

The first factor will be zero when z = 3, so that is one solution.

The second factor will be zero when ...

(z +1.5)² +6.75 = 0 . . . . . . . complete the square

(z +1.5) = ±√(-6.75) = ±√(-27/4) = ±(3/2)√(-3) . . . subtract 6.75; square root

z = -1.5 ±1.5√3i

The three solutions to the equation are ...

• z = 3
• z = -1.5+1.5(√3)i
• z = -1.5-1.5(√3)i

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