Question

add item review: $sqrt3{48m^{11}n^{24}}$, $sqrt3{2cdot6m^{11}n^{24}}$, $sqrt3{2cdot6m^{9 + 2}n^{24}}$, $sqrt3{2^{3}cdot6m^{9+2}n^{24}}$, $sqrt3{2^{3}cdot6m^{9}m^{2}n^{24}}$, $sqrt3{2^{3}}sqrt3{m^{9}}sqrt3{n^{24}}sqrt3{6m^{2}}$, $2sqrt3{m^{9}}sqrt3{n^{24}}sqrt3{6m^{2}}$, $2m^{3}sqrt3{n^{24}}sqrt3{6m^{2}}$, $2m^{3}n^{8}sqrt3{6m^{2}}$

Explanation:

Step1: Prime - factorize 48

$48 = 2^{4}\times3$

Step2: Rewrite the cube - root expression

$\sqrt[3]{48m^{11}n^{24}}=\sqrt[3]{2^{4}\times3\times m^{11}n^{24}}$

Step3: Use the property $\sqrt[3]{abc}=\sqrt[3]{a}\times\sqrt[3]{b}\times\sqrt[3]{c}$

$\sqrt[3]{2^{4}\times3\times m^{11}n^{24}}=\sqrt[3]{2^{4}}\times\sqrt[3]{3}\times\sqrt[3]{m^{11}}\times\sqrt[3]{n^{24}}$

Step4: Simplify each cube - root

$\sqrt[3]{2^{4}} = 2\sqrt[3]{2}$, $\sqrt[3]{m^{11}}=m^{3}\sqrt[3]{m^{2}}$, $\sqrt[3]{n^{24}} = n^{8}$

Step5: Combine the terms

$2m^{3}n^{8}\sqrt[3]{6m^{2}}$

Answer:

$2m^{3}n^{8}\sqrt[3]{6m^{2}}$