Solution

\[ \begin{array}{c|c} 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array} \quad \quad \begin{array}{c|c} 2 & 288 \\ \hline 2 & 144 \\ \hline 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array} \] \[ \begin{align*} & = \sqrt[3]{288 \ \sqrt[3]{72 \ \sqrt[3]{27}}} \\ & = \sqrt[3]{288 \ \sqrt[3]{72 \ \sqrt[3]{ \boxed{3 \times 3 \times 3}}}} \\ & = \sqrt[3]{288 \ \sqrt[3]{72 \times 3}} \\ & = \sqrt[3]{288 \ \sqrt[3]{\boxed{2 \times 2 \times 2} \times \boxed{3 \times 3 \times 3}}} \\ & = \sqrt[3]{288 \times 2 \times 3} \\ & = \sqrt[3]{\boxed{2 \times 2 \times 2} \times \boxed{2 \times 2 \times 2} \times \boxed{3 \times 3 \times 3}} \\ & = 2 \times 2 \times 3 \\ & = 12 \\ \end{align*} \]

Answer \( \sqrt[3]{288 \ \sqrt[3]{72 \ \sqrt[3]{27}}} = \color{red} 12 \)

Solution

\[ \begin{align*} \\ \text{Let the numbers be } & 2x, 3x, 4x \\ (2x)^3 + (3x)^3 + (4x)^3 & = 33957 \\ 8x^3 + 27x^3 + 64x^3 & = 33957 \\ 99x^3 & = 33957 \\ x^3 & = \frac{\cancel{33957}^{343}}{\cancel{99}_1} \\ x^3 & = 343 \\ x & = \sqrt[3]{343} \\ x & = \sqrt[3]{7 \times 7 \times 7} \\ x & = 7 \\ \\ \text{The numbers are } \implies 2x &= 14 \\ \implies 3x &= 21 \\ \implies 4x &= 28 \\ \end{align*} \]

Answer The numbers are \( = \color{red} 14, 21, 28 \)