Question

Among the solutions of the equation \( x^6 = 64 \), the sum of all those solutions whose real part is negative is:

• A. \( 0 \)
• B. \( -2 \)
• C. \( -4 \)
• D. \( -3 \)

Hint:

For \( x^n = a \), the roots are symmetric about the origin. The sum of all roots is 0. Subtract the roots with positive real parts from the total sum (which is 0) to find the sum of negative real parts.

Answer 1

The Correct Option is C

Step 1: Set up the sixth roots.
We solve $x^6 = 64 = 2^6$. The six roots all have modulus $2$ and are equally spaced by $60^\circ$ around the circle of radius $2$.
Step 2: Write the roots.
They are $x_k = 2\big(\cos\frac{k\pi}{3} + i\sin\frac{k\pi}{3}\big)$ for $k = 0,1,2,3,4,5$.
Step 3: Compute their real parts.
The real parts are $2\cos\frac{k\pi}{3}$: for $k=0,1,2,3,4,5$ these are $2,\ 1,\ -1,\ -2,\ -1,\ 1$.
Step 4: Keep only negative real parts.
The roots with negative real part are at $k=2,3,4$: \[ -1 + i\sqrt{3},\quad -2,\quad -1 - i\sqrt{3}. \]
Step 5: Add them.
The imaginary parts $+\sqrt{3}$ and $-\sqrt{3}$ cancel, leaving the real sum $(-1) + (-2) + (-1) = -4$.
Step 6: State the result.
So the sum is $-4$, which is option (C).
\[ \boxed{-4} \]