\sqrt[3]{x} \left(x^{2} - 16\right) when x = -4/root(7)
You asked:
Calculate the value of \(\sqrt[3]{x} \cdot \left({x}^{2} - 16\right)\) when \(x = \frac{-4}{\sqrt[2]{7}}\).
MathBot Answer:
\[\sqrt[3]{x} \cdot \left({x}^{2} - 16\right) = \sqrt[3]{\frac{-4}{\sqrt[2]{7}}} \cdot \left({\left( \frac{-4}{\sqrt[2]{7}} \right)}^{2} - 16\right) = - \frac{96 \sqrt[3]{-1} \cdot 2^{\frac{2}{3}} \cdot 7^{\frac{5}{6}}}{49}\]