Cube Roots of Cubes of Cubes

Relevant wiki: Rules of Exponents

$\sqrt[3]{x}$ can be expressed as $x^{\frac{1}{3}}$ . Thus, the expression becomes $\large \color{#20A900}(3^{\color{#3D99F6}{3}^{\color{#D61F06}{3}}})^{\color{#69047E}\frac{1}{3}}$ which equals $\large \color{#20A900}(3^{\color{#EC7300}{27}})^{\color{#69047E}\frac{1}{3}}$ which equals $\large \color{#20A900}3^{\color{grey}{9}}$ or $\large \boxed{\color{#20A900}3^{\color{#3D99F6}{3}^{\color{#D61F06}{2}}}}$

$\large {\color{#E81990}\sqrt[3]{{\color{#333333}3}^{\color{#3D99F6}3^3}}} = 3^{{\color{#3D99F6}3^3} \times {\color{#E81990}3^{-1}}} = \boxed{\color{#20A900}3^{3^2}}$

Note that in general we can write $\large \color{#E81990} \sqrt[n]{\color{#3D99F6}x}$ as $\large {\color{#3D99F6}x}^{\color{#E81990}\frac{1}{n}} = {\color{#3D99F6}x}^{\color{#E81990}n^{-1}}$

Relevant wiki: Rules of Exponents

Relevant wiki: Rules of Exponents

$\sqrt[3]{3^{3^{3}}}=3^{\left(3^3 \cdot \frac{1}{3}\right)}=3^{3^3\cdot 3^{-1}}$ $3^{3^3\cdot 3^{-1}}=3^{3^{3-1}}=\boxed{3^{3^{2}}}$

(3^27)^(1/3)=3^9

We're simple looking for the cube root of 3 raised to the power 27.

That's 3^9, ie 3 raised to 3 squared.

Note that:

1. $x^n × x^m = x^{n+m}$
2. $(x^n)^m = x^{n×m}$
3. $x^{n^m} = x^{(n^m)}$
4. $\sqrt[n]{x} = x^{\frac 1n}$

So we have $\sqrt[3]{3^{3^3}}$ which can be rewritten (using rules 3 & 4 above. 1 was only included to give an idea of how exponents work. Note how rule 1 turns multiplication into addition and is the basis for logarithms) $(3^{27})^{\frac 13}$ and using rule 2 that gives us $3^9 = 3^{3^2}$