The sum of all real values of $$ k $$ for which

$$
\left(\frac{1}{8} ight)^k imes\left(\frac{1}{32768} ight)^{\frac{1}{3}}=\frac{1}{8} imes\left(\frac{1}{32768} ight)^{\frac{1}{k}}
$$
, is

Question

The sum of all real values of $$ k $$ for which

$$
\left(\frac{1}{8}ight)^k	imes\left(\frac{1}{32768}ight)^{\frac{1}{3}}=\frac{1}{8}	imes\left(\frac{1}{32768}ight)^{\frac{1}{k}}
$$
, is

Accepted Answer

The problem requires finding the sum of all real values of $$k$$ that satisfy the given equation. Here's the detailed solution:

**Given Equation:**

$$
   \left(\frac{1}{8}ight)^k 	imes \left(\frac{1}{32768}ight)^{\frac{1}{3}} = \frac{1}{8} 	imes \left(\frac{1}{32768}ight)^{\frac{1}{k}}
   $$

**Simplify $$32768$$:**

Recognize that:
   $$
   32768 = 8^5
   $$
   Substituting this into the equation:
   $$
   \left(\frac{1}{8}ight)^k 	imes \left(\frac{1}{8}ight)^{5 \cdot \frac{1}{3}} = \left(\frac{1}{8}ight) 	imes \left(\frac{1}{8}ight)^{5 \cdot \frac{1}{k}}
   $$

**Combine Powers:**

Since the bases are equal ($$1/8$$), equate the exponents:
   $$
   k + \frac{5}{3} = 1 + \frac{5}{k}
   $$

**Rearrange and Simplify:**

Multiply through by $$k$$ to eliminate the fraction:
   $$
   k^2 + \frac{5k}{3} = k + 5
   $$
   Multiply through by 3 to clear the denominator:
   $$
   3k^2 + 5k = 3k + 15
   $$
   Simplify:
   $$
   3k^2 + 2k - 15 = 0
   $$

**Solve Quadratic Equation:**

The quadratic equation is:
   $$
   3k^2 + 2k - 15 = 0
   $$
   The sum of the roots of a quadratic equation $$ax^2 + bx + c = 0$$ is given by:
   $$
   -\frac{b}{a}
   $$
   Here, $$a = 3$$, $$b = 2$$, and $$c = -15$$. So:
   $$
   -\frac{b}{a} = -\frac{2}{3}
   $$

**Final Answer:**

The sum of all real values of $$k$$ is **$$-\frac{2}{3}$$**.

### Final Answer:
The correct choice is
 $$ -\frac{2}{3} $$ .

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