Bringing terms under a common radical sign (review II)

Text

Bring under a single radical sign:

• (a) $\sqrt[6]{3^4}\cdot\sqrt[8]{5^3}$;
• (b) $\sqrt[12]{5^3}\cdot\sqrt[18]{5^3}\cdot\sqrt[15]{7^3}$;
• (c) $\sqrt[3]{3^2}\cdot\sqrt[4]{2^3}\cdot\sqrt[5]{7^4}$.

Solution

Bring each factor to the common index (l.c.m. of the indices) and collect under a single radical.

(a) l.c.m.$(6,8)=24$: $\sqrt[6]{3^4}=\sqrt[24]{3^{16}}$ and $\sqrt[8]{5^3}=\sqrt[24]{5^9}$, hence $\sqrt[6]{3^4}\cdot\sqrt[8]{5^3}=\sqrt[24]{3^{16}\cdot5^9}.$

(b) Simplify first: $\sqrt[12]{5^3}=\sqrt[4]{5}=5^{1/4}$, $\sqrt[18]{5^3}=\sqrt[6]{5}=5^{1/6}$, $\sqrt[15]{7^3}=\sqrt[5]{7}=7^{1/5}$. The exponent of $5$ is $\tfrac14+\tfrac16=\tfrac{5}{12}$; l.c.m.$(12,5)=60$: $\sqrt[12]{5^3}\cdot\sqrt[18]{5^3}\cdot\sqrt[15]{7^3}=5^{5/12}\,7^{1/5}=\sqrt[60]{5^{25}\cdot7^{12}}.$

(c) l.c.m.$(3,4,5)=60$: $\sqrt[3]{3^2}=\sqrt[60]{3^{40}}$, $\sqrt[4]{2^3}=\sqrt[60]{2^{45}}$, $\sqrt[5]{7^4}=\sqrt[60]{7^{48}}$, hence $\sqrt[3]{3^2}\cdot\sqrt[4]{2^3}\cdot\sqrt[5]{7^4}=\sqrt[60]{2^{45}\cdot3^{40}\cdot7^{48}}.$

$\boxed{(a)\ \sqrt[24]{3^{16}\cdot5^9}\qquad (b)\ \sqrt[60]{5^{25}\cdot7^{12}}\qquad (c)\ \sqrt[60]{2^{45}\cdot3^{40}\cdot7^{48}}}$