My Attempt:
$t$ values that satisfy $\frac{\sqrt{3+\sqrt[3]{t}}}{\sqrt{3-\sqrt[3]{t}}} = 3$
Move the denominator to the right side of equality via multiplication.
$\sqrt{3+\sqrt[3]{t}} = 3\sqrt{3-\sqrt[3]{t}}$
Then
$\left(\sqrt{3+\sqrt[3]{t}}\right)^{2} = \left(3\sqrt{3-\sqrt[3]{t}}\right)^{2}$
Then
$3+\sqrt[3]{t} = 9\left(3-\sqrt[3]{t}\right) \therefore 3+\sqrt[3]{t} = 27-3\sqrt[3]{t}$
Move the constants to the right side.
$\sqrt[3]{t} = 24 - 3\sqrt[3]{t}$
Move the radical to the left
$4\sqrt[3]{t} = 24$
Removing the $4$ coefficient.
$\sqrt[3]{t} = 6$
Getting rid of the radical.
$\left(\sqrt[3]{t}\right)^{3} = 6^{3} \therefore t = 216$
When plugging this number back into the original equation, I get a denominator with $\sqrt{-3}$ which makes no sense.