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)

2. Originally Posted by ccdalamp
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)
$x^3=-1= cis((2n+1) \pi )\ \ \ n=0,1,2, ...$

so

$
x=cis\left(\frac{2n+1}{3}\pi \right)\ \ \ n=0,1,2, ...
$

which if I am not mistaken is answer (c)

RonL

3. 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: . $x^3 + 1\:=\:0$

$a)\;\text{-}1,\:1,\:\text{cis}\left(\frac{\pi}{3}\right)\qqua d\qquad\qquad b)\;1,\:\text{cis}\left(\frac{\pi}{3}\right),\:\te xt{cis}\left(\frac{2\pi}{3}\right)$
$c)\;\text{-}1,\:\text{cis}\left(\frac{\pi}{3}\right),\:\text{ cis}\left(\frac{5\pi}{3}\right)\qquad d)\;\text{cis}\left(\frac{\pi}{3}\right),\:\text{c is}\left(\frac{2\pi}{3}\right),\:\text{cis}\left(\ frac{4\pi}{3}\right)$

Factor: . $(x + 1)(x^2 - x + 1) \:=\:0$

And we have: . $\begin{array}{ccccccc}x + 1 & = & 0 & \;\;\Rightarrow\;\; & x & = & -1 \\
x^2-x+1 & = & 0 & \Rightarrow & x & = & \frac{1\pm i\sqrt{3}}{2} \end{array}$

And the complex roots can be rewritten: . $\begin{array}{ccc}\frac{1}{2} + \frac{\sqrt{3}}{2}i & \Rightarrow & \cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right) \\
\frac{1}{2} - \frac{\sqrt{3}}{2}i & \Rightarrow & \cos\left(\frac{5\pi}{3}\right) + i\sin\left(\frac{5\pi}{3}\right) \end{array}$

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

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: . $\sqrt[3]{\text{-}1} = \text{-}1$

Therefore, (c) is the only candidate.