Volume of a cube in a different way

The longest rod that can be fit inside a cube is its diagnol which can be found by the formula $a\sqrt{3}$ for a cube of side length $a$ .

$\begin{aligned} a\sqrt{3} & = \sqrt[6]{2187}\\ a\sqrt{3} & = \sqrt[6]{729 × 3}\\ a & = \dfrac{3 × \sqrt[6]{3}}{\sqrt{3}}\\ & = \sqrt{3} × \sqrt[6]{3}\\ & = 3^{\tfrac{1}{2}} × 3^{\tfrac{1}{6}}\\ & = 3^{\tfrac{2}{3}}\\ \text{Volume} & = a^3\\ a^3 = {\left(3^{\tfrac{2}{3}}\right)}^3\\ & = 3^2\\ & = \boxed{9.00} \end{aligned}$

Lol, I'd found a similar question in my 8th standard maths book.

It is evident that the longest possible one-dimensional rod that can be fitted inside a cube is equal to the length of its diagonal.

Let the side of the cube be $\displaystyle a$ , then its diagonal will be $\displaystyle \sqrt{3}a$ , then:

$\displaystyle diagonal = (2187)^{\frac{1}{6}}$

$\displaystyle \implies \sqrt{3}a = (2187)^{\frac{1}{6}}$

Cubing both sides, we have:

$\displaystyle \implies 3 \sqrt{3}a^{3} = [(2187)^{\frac{1}{6}}]^{3}$

$\displaystyle \implies 3 \sqrt{3}a^{3} = (2187)^{\frac{1}{2}}$

$\displaystyle \implies 3 \sqrt{3}a^{3} = \sqrt{2187}$

$\displaystyle \implies a^{3} = \frac{\sqrt{2187}}{3 \sqrt{3}}$

$\displaystyle \implies a^{3} = \frac{1}{3} \times \frac{\sqrt{2187}}{\sqrt{3}}$

$\displaystyle \implies a^{3} = \frac{1}{3} \times \sqrt{\frac{2187}{3}}$

$\displaystyle \implies a^{3} = \frac{1}{3} \times \sqrt{729}$

$\displaystyle \implies a^{3} = \frac{1}{3} \times 27$

$\displaystyle \implies a^{3} = 9$ $\displaystyle \implies \boxed{V=9}$