First, we re-write the expression as a square: \[ \sqrt{6} \cdot \sqrt{6} = \left(\sqrt{6}\right)^2. \] The square and the square root cancel each other out. So, we have \[ \left(\sqrt{6}\right)^2 = 6. \]

First, we re-write the expression as a square: \[ \sqrt{7} \cdot \sqrt{7} = \left(\sqrt{7}\right)^2. \] The square and the square root cancel each other out. So, we have \[ \left(\sqrt{7}\right)^2 = 7. \]

First, we re-write the product as a cube: \[ \sqrt[3]{-11} \cdot \sqrt[3]{-11} \cdot \sqrt[3]{-11} = \left(\sqrt[3]{-11}\right)^3. \] The cube and the cube root cancel each other out. So, we have \[ \left(\sqrt[3]{-11}\right)^3 = -11. \]

First, we re-write the expression as the product of a cube and a square:

\begin{align*} {\color{blue}\sqrt[3]{-5}} \cdot {\color{red}\sqrt{7}} \cdot {\color{blue}\sqrt[3]{-5}} \cdot {\color{red}\sqrt{7}} \cdot {\color{blue}\sqrt[3]{-5}} &= \\[5pt] \left({\color{blue}\sqrt[3]{-5}} \cdot {\color{blue}\sqrt[3]{-5}} \cdot {\color{blue}\sqrt[3]{-5}}\right) \left({\color{red}\sqrt{7}} \cdot {\color{red}\sqrt{7}}\right) &= \\[5pt] \left({\color{blue}\sqrt[3]{-5}}\right)^3 \left({\color{red}\sqrt{7}}\right)^2 &= \\ \end{align*}

The cube and the cube root cancel each other out. Similarly, the square and the square root also cancel. So, we have

\begin{align*} \left({\color{blue}\sqrt[3]{-5}}\right)^3 \left({\color{red}\sqrt{7}}\right)^2 &= \\[5pt] {\color{blue}(-5)} \cdot {\color{red}7} &= \\[5pt] -35&. \end{align*}