Simplifying radicals (extra practice)

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Simplify as much as possible:

• (a) $\sqrt{3^3\cdot\sqrt[5]{3}\cdot\sqrt{5}}$
• (b) $\sqrt{12}\,\sqrt[3]{36}\,\sqrt[5]{24}$
• (c) $\sqrt{2\cdot\sqrt[4]{2\cdot\sqrt[3]{2}}}$
• (d) $\dfrac{\sqrt[7]{5^2}\,\sqrt[3]{5^2}}{\sqrt{5^3}}$

Solution

(a) Inside: $3^{3}\cdot 3^{1/5}\cdot 5^{1/2}=3^{16/5}\cdot 5^{1/2}$; square root: $3^{8/5}\cdot 5^{1/4}$. Common index $20$: $\sqrt[20]{3^{32}\cdot 5^{5}}=3\sqrt[20]{3^{12}\cdot 5^{5}}$.

(b) $12=2^2\cdot 3$, $36=2^2\cdot 3^2$, $24=2^3\cdot 3$; common index $30$: $\sqrt{12}\,\sqrt[3]{36}=\sqrt[30]{2^{50}\cdot 3^{35}}$ and $\sqrt[5]{24}=\sqrt[30]{2^{18}\cdot 3^{6}}$, hence $\sqrt[30]{2^{68}\cdot 3^{41}}=2^2\cdot 3\sqrt[30]{2^{8}\cdot 3^{11}}=12\sqrt[30]{2^{8}\cdot 3^{11}}$.

(c) Nested radicals: $\sqrt[3]2=2^{1/3}$, $2\cdot 2^{1/3}=2^{4/3}$, $\sqrt[4]{2^{4/3}}=2^{1/3}$, $2\cdot 2^{1/3}=2^{4/3}$, $\sqrt{2^{4/3}}=2^{2/3}=\sqrt[3]{4}$.

(d) $5^{2/7+2/3-3/2}=5^{(12+28-63)/42}=5^{-23/42}=\dfrac{1}{\sqrt[42]{5^{23}}}$.

$\boxed{\;(a)\;3\sqrt[20]{3^{12}\cdot 5^{5}}\qquad (b)\;12\sqrt[30]{2^{8}\cdot 3^{11}}\qquad (c)\;\sqrt[3]{4}\qquad (d)\;\dfrac{1}{\sqrt[42]{5^{23}}}\;}$