Solve.
-$x ^ { 3 } - 125 = 0$
A) $5 , \frac { - 5 - 5 i \sqrt { 3 } } { 2 } , \frac { - 5 + 5 i \sqrt { 3 } } { 2 }$
B) $- 5 , \frac { 5 - 5 i \sqrt { 3 } } { 2 } , \frac { 5 + 5 i \sqrt { 3 } } { 2 }$
C) $5 , \frac { 5 - 5 i \sqrt { 3 } } { 2 } , \frac { 5 + 5 i \sqrt { 3 } } { 2 }$
D) $- 5 , \frac { 5 - 5 \sqrt { 5 } } { 2 } , \frac { 5 + 5 \sqrt { 5 } } { 2 }$

Question

Quizplus · Accepted Answer

The equation $x^3 - 125 = 0$ can be solved by recognizing it as a difference of cubes, $x^3 - 5^3 = 0$, which factors to $(x-5)(x^2+5x+25)=0$. The real root is $x=5$, and the complex roots are obtained by solving the quadratic equation, yielding $\frac{-5 \pm i\sqrt{75}}{2}$, which simplifies to $\frac{-5 \pm 5i\sqrt{3}}{2}$.

Solve - x3−125=0x ^ { 3 } - 125 = 0x3−125=0

Solve
- $x ^ { 3 } - 125 = 0$