Question

4³√27*³√5+³√8*³√5

Answer 1

you first take common \[\sqrt[3]{5}\]
so the remaining value is \[4\sqrt[3]{27}+\sqrt[3]{8}\]
now 27 is the cube root of 3 so \[4\left( 3^{3} \right)^{1\3}\] 8 is also the cube root of 2 we can write it as\[\left( 2^3 \right)^{1\3}\]
now, power of both values cancel then the remaining value will be
\[\sqrt[3]{5}\left\{4(3)+(2) \right\}\]
\[\sqrt[3]{5}\left\{ 12+2 \right\}\]
\[\sqrt[3]{5}\left( 14 \right)\] adn the answer is
\[14\sqrt[3]{5}\] or this \[14\left( 5^{1\3} \right)\]