Find the realand complex solutions to the equation X3(cubed) + 1=0

my choices

a. -1,1,cis(pi/3)

b. 1, cis(pi/3), cis (2pi/3)

c. -1, cis(pi/3), cis(5pi/3)

d. cis (pi/3), cis(2pi/3), cis(4pi/3)

Results 1 to 3 of 3

- July 10th 2007, 12:27 PM #1

- Joined
- Jul 2007
- Posts
- 8

## please help with this

Find the realand complex solutions to the equation X3(cubed) + 1=0

my choices

a. -1,1,cis(pi/3)

b. 1, cis(pi/3), cis (2pi/3)

c. -1, cis(pi/3), cis(5pi/3)

d. cis (pi/3), cis(2pi/3), cis(4pi/3)

- July 10th 2007, 01:03 PM #2

- Joined
- Nov 2005
- From
- someplace
- Posts
- 14,972
- Thanks
- 4

- July 10th 2007, 04:08 PM #3

- Joined
- May 2006
- From
- Lexington, MA (USA)
- Posts
- 11,981
- Thanks
- 812

Hello, ccdalamp!

With "cis" in the answer choices, I assume that DeMoivre's Formula is expected.

But it can be solved algebraically . . . with some trig.

Find the real and complex solutions to the equation: .

Factor: .

And we have: .

And the complex roots can be rewritten: .

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

On the other hand, there is an "eyeball" solution.

A cubic equation has three roots, of course.

Either all three are real or there is one real and two complex conjugates.

. . Only (c) meets this requirement.

Moreover, we already know that: .

Therefore, (c) is the only candidate.