Question

$$(2^{6}\times3^{4}\times5^{2})^{\frac{3}{2}}; \sqrt[4]{256}\times\sqrt{729}\times\sqrt[3]{125}$$

Answer

$$129600^(3/2);4^(1/256)*27*3^(1/125)$$

Solution


Simplify  \({2}^{6}\)  to  \(64\).
\[\begin{aligned}&{(64\times {3}^{4}\times {5}^{2})}^{\frac{3}{2}}\\&\sqrt[256]{4}\sqrt{729}\sqrt[125]{3}\end{aligned}\]
Simplify  \({3}^{4}\)  to  \(81\).
\[\begin{aligned}&{(64\times 81\times {5}^{2})}^{\frac{3}{2}}\\&\sqrt[256]{4}\sqrt{729}\sqrt[125]{3}\end{aligned}\]
Simplify  \({5}^{2}\)  to  \(25\).
\[\begin{aligned}&{(64\times 81\times 25)}^{\frac{3}{2}}\\&\sqrt[256]{4}\sqrt{729}\sqrt[125]{3}\end{aligned}\]
Simplify  \(64\times 81\)  to  \(5184\).
\[\begin{aligned}&{(5184\times 25)}^{\frac{3}{2}}\\&\sqrt[256]{4}\sqrt{729}\sqrt[125]{3}\end{aligned}\]
Simplify  \(5184\times 25\)  to  \(129600\).
\[\begin{aligned}&{129600}^{\frac{3}{2}}\\&\sqrt[256]{4}\sqrt{729}\sqrt[125]{3}\end{aligned}\]
Since \(27\times 27=729\), the square root of \(729\) is \(27\).
\[\begin{aligned}&{129600}^{\frac{3}{2}}\\&\sqrt[256]{4}\times 27\sqrt[125]{3}\end{aligned}\]