How to quickly solve $ 2 + \frac94t = \left( \frac{27+16\sqrt{2}}{8} \right)^{\frac23} $?

Question

What is a quick and easy method to solve the equation $$ 2 + \frac94t = \left( \frac{27+16\sqrt{2}}{8} \right)^{\frac23} = \frac{\left( 16 \sqrt{2} + 27 \right)^\frac{2}{3}}{4} $$? I don't know how to find a cube root, nor do I have time to do so. This is part of a much bigger question which is supposed to be done in around $4-5$ minutes, and I can spend at the most $1 or 2$ minutes of this. How can I solve this quickly? The answer I'm supposed to get is 0.612 seconds.

Answer 1

The simplest form I can see is $$t=\frac 19\sqrt[3]{1241+864\sqrt{2}}-\frac 89$$

Done by evaluating $8^{\frac 23}$, and $(27+16\sqrt2)^2$, then rearranging for $t$.

From this we get $t\approx0.6116201124$. We can achieve something close by saying $\sqrt2\approx\ 1.5$, then $t\approx\frac 19(\sqrt[3]{2537}-8)\approx\frac19(13.5-8)=\frac 19(5.5)=\frac19\cdot\frac{11}{2}=\frac{11}{18}$

Hence we get $t\approx0.6\dot{1}$

Answer 2

Multiply by $t$ and use the quadratic formula.

$$t^2\bigg(\frac 94\bigg) - t\bigg(\frac{(27+16\sqrt{2})^{2/3}}{4}-2\bigg)+0=0.$$

Let $u=-\bigg(\dfrac{(27+16\sqrt{2})^{2/3}}{4}-2\bigg)=2-\dfrac{(27+16\sqrt{2})^{2/3}}{4}$ then $$\begin{align}t&=\frac{u\pm \sqrt{u^2-0}}{2(9/4)} \\ &=\frac{2u}{9}\text{ or } 0.\end{align}$$ The $u$'s cancel out or the $2$'s cancel out, and $2/4=1/2$. But trivially, $t\neq 0$.

$$\therefore t=\frac {2u}{9}\tag*{$\bigcirc$}$$