We should start by recognizing that this is a QUADRATIC EQUATION in disguise.
Notice that \(x^{2/3} = (x^{1/3})^{2}\)
So let's let \(x^{1/3}\) = k, and replace \(x^{1/3}\) with k to get: 3k² = 54 + 9k
Rearrange to get: 3k² - 9k - 54 = 0
Factor: 3(k² - 3k - 18) = 0
Factor more: 3(k - 6)(k + 3) = 0
So, the solutions are k = 6 and k = -3

Since k = \(x^{1/3}\), we can write: \(x^{1/3}\) = 6 and \(x^{1/3}\) = -3

If \(x^{1/3}\) = 6, then x = 6³ = 216
If \(x^{1/3}\) = -3, then x = (-3)³ = -27

So, the sum of all solutions = 216 + (-27) = 189

Answer: B

Cheers,
Brent
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What is the sum of all solutions to the equation \(3x^{2/3} = 54 + 9x^{1/3}\)

Let \(x^{1/3} = a\)

\(3a^2 = 54 + 9a\)

\(3a^2 - 9a - 54 = 0\)

\(a^2 - 3a - 18 = 0\)

\((a - 6) (a + 3) = 0\)

\(a = 6\) Or

\(a = -3\)


As, \(x^{1/3} = a\)

\(x^{1/3} = 6\)

\(x = 6 * 6 * 6 = 216\)

\(x^{1/3} = -3\)

\(x = -3 * -3 * -3 = -27\)

So, sum of the roots \(= 216 + (-27) = 216 - 27 = 189\)


Hence, Answer is B

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