Simplifying radicals (review II)

Text

Simplify as much as possible:

• (a) $\sqrt{7^{24}\cdot\sqrt[15]{7}\cdot\sqrt[12]{7}}$;
• (b) $\sqrt{256}\,\sqrt[4]{72}\,\sqrt[7]{48}$;
• (c) $\sqrt{5^7\cdot\sqrt[4]{5^3\cdot\sqrt[3]{5^8}}}$;
• (d) $\dfrac{\sqrt[7]{3^2}\,\sqrt[3]{5^2}}{\sqrt{5^3}}$.

Solution

It is convenient to write everything with fractional exponents and factor into primes.

(a) Under the radical the exponent of $7$ is $24+\tfrac1{15}+\tfrac1{12}=24+\tfrac{3}{20}=\tfrac{483}{20}$; the square root halves it: $7^{483/40}$. Since $\tfrac{483}{40}=12+\tfrac{3}{40}$, $\sqrt{7^{24}\cdot\sqrt[15]{7}\cdot\sqrt[12]{7}}=7^{12}\sqrt[40]{7^3}=7^{12}\sqrt[40]{343}.$

(b) $256=2^8$, $72=2^3\cdot3^2$, $48=2^4\cdot3$, so $\sqrt{256}=2^4$, $\sqrt[4]{72}=2^{3/4}3^{1/2}$, $\sqrt[7]{48}=2^{4/7}3^{1/7}$. The exponent of $2$ is $4+\tfrac34+\tfrac47=5+\tfrac{9}{28}$, that of $3$ is $\tfrac12+\tfrac17=\tfrac{9}{14}$: $\sqrt{256}\,\sqrt[4]{72}\,\sqrt[7]{48}=2^5\cdot2^{9/28}\cdot3^{9/14}=32\sqrt[28]{2^9\cdot3^{18}}.$

(c) From the inside: $\sqrt[3]{5^8}=5^{8/3}$, then $5^3\cdot5^{8/3}=5^{17/3}$, then $\sqrt[4]{5^{17/3}}=5^{17/12}$, then $5^7\cdot5^{17/12}=5^{101/12}$, finally the square root: $5^{101/24}$. Since $\tfrac{101}{24}=4+\tfrac{5}{24}$, $\sqrt{5^7\cdot\sqrt[4]{5^3\cdot\sqrt[3]{5^8}}}=5^{4}\sqrt[24]{5^5}=625\sqrt[24]{3125}.$

(d) The exponent of $3$ is $\tfrac27$, that of $5$ is $\tfrac23-\tfrac32=-\tfrac56$: $\frac{\sqrt[7]{3^2}\,\sqrt[3]{5^2}}{\sqrt{5^3}}=3^{2/7}\,5^{-5/6}=\frac{\sqrt[7]{9}}{\sqrt[6]{5^5}}=\frac{\sqrt[7]{9}\,\sqrt[6]{5}}{5}.$

$\boxed{(a)\ 7^{12}\sqrt[40]{343}\qquad (b)\ 32\sqrt[28]{2^9\cdot3^{18}}\qquad (c)\ 625\sqrt[24]{3125}\qquad (d)\ \dfrac{\sqrt[7]{9}\,\sqrt[6]{5}}{5}}$