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

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



You asked:

Evaluate the expression: (27+33643)22+15\frac{\left(\sqrt{27 + {3}^{3}} - \sqrt[3]{64}\right) \cdot {2}^{2 + 1}}{5}

MathBot Answer:

(27+33643)22+15=325+24655.35755076535925487134696355858828\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


(27+33643)22+15=(27+33643)22+115=(27+27643)22+115=(54643)22+115=(36643)22+115=(364)22+115=(4+36)22+115=(4+36)2315=(4+36)85=325+2456\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}