ABC26GN1965 · Surds and Indices

$(216)^{\frac{3}{5}} \times(2500)^{\frac{2}{5}} \times(300)^{\frac{1}{5}}=\left(3^{3} \times 2^{3}\right)^{\frac{3}{5}} \times\left(5^{4} \times 2^{2}\right)^{\frac{2}{5}}$ $$\begin{aligned} & =3^{\left(3 \times \frac{3}{5}\right)} \times 2^{\left(3 \times \frac{3}{5}\right)} \times 5^{\left(4 \times \frac{2}{5}\right)} \times 2^{\left(2 \times \frac{2}{5}\right)} \times 5^{\left(2 \times \frac{1}{5}\right)} \\ & \times 2^{\left(2 \times \frac{1}{5}\right)} \times 3^{\frac{1}{5}} \\ & =3^{\frac{9}{5}} \times 2^{\frac{9}{5}} \times 5^{\frac{8}{5}} \times 2^{\frac{4}{5}} \times 5^{\frac{2}{5}} \times 2^{\frac{2}{5}} \times 3^{\frac{1}{5}} \\ & =3^{\left(\frac{9}{5}+\frac{1}{5}\right)} \times 2^{\left(\frac{9}{5}+\frac{4}{5}+\frac{2}{5}\right)} \times 5^{\left(\frac{8}{5}+\frac{2}{5}\right)}=3^{2} \times 2^{3} \times 5^{2} \end{aligned}$$ Hence, the number of prime factors $=(2+3+2)=7$.