Question
State which of the following are surds Justify. : $\sqrt[3]{64}$
✓
Answer
$\sqrt[3]{64}$ is not a surd because
$
\begin{aligned}
\sqrt[3]{64} & =(64)^{\frac{1}{3}} \\
& =\left(4^3\right)^{\frac{1}{3}}
\end{aligned}
$
$=4$, which is not an irrational number.
$
\begin{aligned}
\sqrt[3]{64} & =(64)^{\frac{1}{3}} \\
& =\left(4^3\right)^{\frac{1}{3}}
\end{aligned}
$
$=4$, which is not an irrational number.
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