Approach #1
If we recognize that $x^{\frac{2}{3}} = (x^{\frac{1}{3}})^2$, then we can use a technique called u-substitution.
Let $u = x^{\frac{1}{3}}$

Now take original equation and replace $x^{\frac{1}{3}}$ with $u$ to get: u² - u - 2 = 4
Subtract 4 from both sides to get: to get: u² - u - 6 = 0
Factor: (u - 3)(u + 2) = 0
So, the two solutions here are u = 3 and u = -2

At this point, we replace u with $x^{\frac{1}{3}}$ to get:
$x^{\frac{1}{3}}$ = 3 and $x^{\frac{1}{3}}$ = -2

If $x^{\frac{1}{3}}$ = 3, then x = 27
If $x^{\frac{1}{3}}$ = -2, then x = -8

So, the SUM of the solutions = 27 + -8 = 19

Answer: D
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Approach #2
If we recognize that $x^{\frac{2}{3}} = (x^{\frac{1}{3}})^2$, then we can go straight to factoring.

GIVEN: $x^{\frac{2}{3}} - x^{\frac{1}{3}} - 2 = 4$
Subtract 4 from both sides to get: $x^{\frac{2}{3}} - x^{\frac{1}{3}} - 6 = 0$
Factor: $(x^{\frac{1}{3}} - 3)(x^{\frac{1}{3}} + 2) = 0$
So, EITHER $x^{\frac{1}{3}}$ = 3 OR $x^{\frac{1}{3}}$ = -2

If $x^{\frac{1}{3}}$ = 3, then x = 27
If $x^{\frac{1}{3}}$ = -2, then x = -8

So, the SUM of the solutions = 27 + -8 = 19

Answer: D

Cheers,
Brent