Solve the cubic equation using factoring and the quadratic formula.
-$$x ^ { 3 } + 1 = 0$$
A) $\{ - 1 , \pm i \}$
B) $\left\{ - 1 , \frac { 1 } { 2 } \pm \frac { \sqrt { 3 } } { 2 } i \right\}$
C) $\{ - 1,1 \pm i \sqrt { 3 } \}$
D) $\{ - 1 , - 1 \pm i \sqrt { 3 } \}$

Question

Quizplus · Accepted Answer

First, rewrite the equation as $x^3 = -1$. Recognizing that $-1$ can be written as $-1 = e^{i\pi} = \cos(\pi) + i\sin(\pi)$, we can use De Moivre's theorem to find the cube roots of $-1$. The roots are given by $x = \sqrt[3]{-1} = \sqrt[3]{e^{i\pi}} = e^{i(\pi + 2k\pi)/3}$ for $k = 0, 1, 2$. For $k = 0$, we get $x = e^{i\pi/3} = \cos(\pi/3) + i\sin(\pi/3) = \frac{1}{2} + i\frac{\sqrt{3}}{2}$. For $k = 1$, we get $x = e^{i\pi} = -1$. For $k = 2$, we get $x = e^{i5\pi/3} = \cos(5\pi/3) + i\sin(5\pi/3) = \frac{1}{2} - i\frac{\sqrt{3}}{2}$. Thus, the roots are $\left\{ - 1 , \frac { 1 } { 2 } \pm \frac { \sqrt { 3 } } { 2 } i \right\}$.

Solve the Cubic Equation Using Factoring and the Quadratic Formula x3+1=0x ^ { 3 } + 1 = 0x3+1=0

Solve the Cubic Equation Using Factoring and the Quadratic Formula $x ^ { 3 } + 1 = 0$