What are the real or imaginary solutions of the polynomial equation? x3 = 216 (1 point) Responses –6, 3 + 3i, and 3 – 3i –6, 3 + 3 i Image with alt text: square root seven , and 3 – 3 i Image with alt text: square root seven –6, 3 + 3i, and 3 – 3i –6, 3 + 3 i Image with alt text: square root of three , and 3 – 3 i Image with alt text: square root of three 6, 3 + 3i, and 3 – 3i 6, 3 + 3 i Image with alt text: square root seven , and 3 – 3 i Image with alt text: square root seven 6, –3 + 3i, and –3 – 3i 6, –3 + 3 i Image with alt text: square root of three , and –3 – 3 i Image with alt text: square root of three

Question

What are the real or imaginary solutions of the polynomial equation? x3 = 216 (1 point) Responses –6, 3 + 3i, and 3 – 3i –6, 3 + 3 i Image with alt text: square root seven , and 3 – 3 i Image with alt text: square root seven –6, 3 + 3i, and 3 – 3i –6, 3 + 3 i Image with alt text: square root of three , and 3 – 3 i Image with alt text: square root of three 6, 3 + 3i, and 3 – 3i 6, 3 + 3 i Image with alt text: square root seven , and 3 – 3 i Image with alt text: square root seven 6, –3 + 3i, and –3 – 3i 6, –3 + 3 i Image with alt text: square root of three , and –3 – 3 i Image with alt text: square root of three

Answer 1

The real and imaginary solutions of the polynomial equation x^3 = 216 are:

-6, 3 + 3i, and 3 - 3i

Answer 2

The real and imaginary solutions of the polynomial equation x^3 = 216 are:

-6, 3 + 3i, and 3 - 3i.

Answer 3

To find the solutions of the polynomial equation x³ = 216, we need to solve for x.

Step 1: Rewrite 216 as a perfect cube.
216 can be written as 6³ since 6 multiplied by itself three times equals 216.

Step 2: Take the cube root of both sides.
Taking the cube root of x³ gives us x.
Taking the cube root of 216 gives us 6.

So, x = 6 is one solution to the equation.

Step 3: Find the other two complex solutions.
To find the other two complex solutions, we use the fact that every complex number has a pair of complex conjugates. The complex conjugate of a + bi is a - bi.

For the equation x³ = 216, subtracting 6 from both sides gives x³ - 6 = 0.

Now, we can factor the left side using the difference of cubes formula:
(a³ - b³) = (a - b)(a² + ab + b²).

In this case, a is x and b is 6. Using the formula, we can factor x³ - 6 as (x - 6)(x² + 6x + 36).

Setting each factor equal to zero gives two additional solutions:
x - 6 = 0, which gives x = 6,
and x² + 6x + 36 = 0.

To solve the quadratic equation x² + 6x + 36 = 0, we can use the quadratic formula:
x = (-b ± √(b² - 4ac)) / (2a).

In this case, a is 1, b is 6, and c is 36. Plugging these values into the quadratic formula gives:
x = (-6 ± √(6² - 4(1)(36))) / (2(1))
x = (-6 ± √(36 - 144)) / 2
x = (-6 ± √(-108)) / 2.

Since the discriminant (√(-108)) is negative, the solutions will be complex numbers.

Simplifying further, we have:
x = (-6 ± i√108) / 2
x = (-6 ± i√(36 * 3)) / 2
x = (-6 ± 6i√3) / 2
x = -3 ± 3i√3.

Therefore, the solutions to the polynomial equation x³ = 216 are:
x = 6,
x = -3 + 3i√3,
and x = -3 - 3i√3.

So, the correct response is: 6, -3 + 3i√3, and -3 - 3i√3.