The sum of all real values of $k$ for which $\small \left( \dfrac{1}{8} \right)^k \times \left( \dfrac{1}{32768} \right)^{\frac{1}{3}} = \left( \dfrac{1}{8} \right) \times \left( \dfrac{1}{32768} \right)^{\frac{1}{k}}$ is

We know that:

(I) $\frac{1}{8} = 2^{-3}$

(II) $\frac{1}{32768} = 2^{-15}$

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Substituting the above and using $a^m \cdot b^m = (ab)^m$:

$(2^{-3})^k \cdot (2^{-15})^{\frac{1}{3}} = ((2^{-3}) \cdot (2^{-15}))^{\frac{1}{k}}$

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Simplifying the left side using the property $(a^m)^n = a^{mn}$:

$(2^{-3})^k \cdot (2^{-15})^{\frac{1}{3}}$

$= 2^{-3k} \cdot 2^{-\frac{15}{3}}$

$= 2^{-3k} \cdot 2^{-5}$

Using the property $a^m \cdot a^n = a^{m+n}$:

$= 2^{-3k-5}$

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Simplifying the right side:

$(2^{-3}) \cdot (2^{-15})^{\frac{1}{k}}$

$= 2^{-3} \cdot 2^{-\frac{15}{k}}$

$= 2^{-3-\frac{15}{k}}$

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Since the bases are equal, we can equate the exponents:

$-3k - 5 = -3 - \dfrac{15}{k}$

$-3k - 5 + 3 + \dfrac{15}{k} = 0$

$-3k - 2 + \dfrac{15}{k} = 0$

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Multiplying throughout by $k$ (assuming $k \neq 0$):

$k(-3k - 2 + \dfrac{15}{k}) = k \cdot 0$

$-3k^2 - 2k + 15 = 0$

$3k^2 + 2k - 15 = 0$

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For any quadratic equation $ax^2 + bx + c = 0$, the sum of roots is given by:

Sum of roots $= \small-\dfrac{b}{a}$

Answer $=\small -\dfrac{2}{3}$