Question
Compare the following:$\sqrt[3]{48}$ and $\sqrt{36}$
✓
Answer
$\sqrt[3]{48}=48^{\frac{1}{3}}$ has power $\frac{1}{3} $
$\sqrt{36}=6$
Now,$ \text{L.C.M.}$ of $3$ and $1=3$
$\sqrt[3]{48}=48^{\frac{1}{3}} $
$\sqrt{36}=6=6^{\frac{3}{3}}=\left(6^3\right)^{\frac{1}{3}}=216^{\frac{1}{3}}$
Since $48<216$,
we have $48^{\frac{1}{3}}<216^{\frac{1}{3}}$
Hence, $\sqrt[3]{48}<\sqrt{36}$.
$\sqrt{36}=6$
Now,$ \text{L.C.M.}$ of $3$ and $1=3$
$\sqrt[3]{48}=48^{\frac{1}{3}} $
$\sqrt{36}=6=6^{\frac{3}{3}}=\left(6^3\right)^{\frac{1}{3}}=216^{\frac{1}{3}}$
Since $48<216$,
we have $48^{\frac{1}{3}}<216^{\frac{1}{3}}$
Hence, $\sqrt[3]{48}<\sqrt{36}$.
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