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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.

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    $\begingroup$ $9(3-\sqrt[3]t)=27-\color{red}9\sqrt[3]t$ $\endgroup$
    – user632577
    Commented Nov 18, 2019 at 5:44

2 Answers 2

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You made a mistake in $$3+\sqrt[3]{t} = 9\left(3-\sqrt[3]{t}\right) \therefore 3+\sqrt[3]{t} = 27-3\sqrt[3]{t}$$ which should have been $$3+\sqrt[3]{t} = 9\left(3-\sqrt[3]{t}\right) \therefore 3+\sqrt[3]{t} = 27-9\sqrt[3]{t}$$

That gives $\sqrt[3]{t} = 2.4$ which satisfies the equation.

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  • $\begingroup$ Hi, I'm still having issue with this. I get $t = 2.4$ and that still doesn't check. $\endgroup$
    – Vocaloidas
    Commented Nov 18, 2019 at 6:01
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    $\begingroup$ @Vocaloidas You are OK. Note that we have $\sqrt[3]{t}=2.4$ thus $\frac{\sqrt{3+\sqrt[3]{t}}}{\sqrt{3-\sqrt[3]{t}}} = \sqrt {5.4/0.6}=\sqrt {9} =3$ $\endgroup$
    – Mohammad Riazi-Kermani
    Commented Nov 18, 2019 at 11:53
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Try a simpler way:

squaring both sides

$$\dfrac91=\dfrac{3+\sqrt[3]t}{3-\sqrt[3]t}$$

Apply Componendo et Dividendo

$$\dfrac{9+1}{9-1}=\dfrac3{\sqrt[3]t}$$

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