(√(27 + 3^3) - ∛(64)) * (2^(2 + 1)) / 5

asked by guest
on Nov 23, 2024 at 1:30 pm



You asked:

Evaluate the expression: \(\frac{\left(\sqrt{27 + {3}^{3}} - \sqrt[3]{64}\right) \cdot {2}^{2 + 1}}{5}\)

MathBot Answer:

\[\frac{\left(\sqrt{27 + {3}^{3}} - \sqrt[3]{64}\right) \cdot {2}^{2 + 1}}{5} = - \frac{32}{5} + \frac{24 \sqrt{6}}{5} \approx 5.35755076535925487134696355858828\]

\[\begin{aligned}\frac{\left(\sqrt{27 + {3}^{3}} - \sqrt[3]{64}\right) \cdot {2}^{2 + 1}}{5}& = \left(\sqrt{27 + 3^{3}} - \sqrt[3]{64}\right) \cdot 2^{2 + 1} \cdot \frac{1}{5}\\& = \left(\sqrt{27 + 27} - \sqrt[3]{64}\right) \cdot 2^{2 + 1} \cdot \frac{1}{5}\\& = \left(\sqrt{54} - \sqrt[3]{64}\right) \cdot 2^{2 + 1} \cdot \frac{1}{5}\\& = \left(3 \cdot \sqrt{6} - \sqrt[3]{64}\right) \cdot 2^{2 + 1} \cdot \frac{1}{5}\\& = \left(3 \cdot \sqrt{6} - 4\right) \cdot 2^{2 + 1} \cdot \frac{1}{5}\\& = \left(-4 + 3 \cdot \sqrt{6}\right) \cdot 2^{2 + 1} \cdot \frac{1}{5}\\& = \left(-4 + 3 \cdot \sqrt{6}\right) \cdot 2^{3} \cdot \frac{1}{5}\\& = \left(-4 + 3 \cdot \sqrt{6}\right) \cdot \frac{8}{5}\\& = - \frac{32}{5} + \frac{24}{5} \cdot \sqrt{6}\end{aligned}\]

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