Determine the total number of distinct prime factors in the expression:
$$
\left(216^{\frac{3}{5}} imes 2500^{\frac{2}{5}} imes 300^{\frac{1}{5}} ight)^5
$$

Question

Determine the total number of distinct prime factors in the expression:  
$$
\left(216^{\frac{3}{5}} 	imes 2500^{\frac{2}{5}} 	imes 300^{\frac{1}{5}}ight)^5
$$

Accepted Answer

To find the number of distinct prime factors, we first simplify the given expression:

1. **Prime Factorization**:  
   - $$216 = 3^3 	imes 2^3$$  
   - $$2500 = 5^4 	imes 2^2$$  
   - $$300 = 3 	imes 2^2 	imes 5^2$$

Substituting these into the expression:  
   $$
   \left(216^{\frac{3}{5}} 	imes 2500^{\frac{2}{5}} 	imes 300^{\frac{1}{5}}ight)^5 = (3^3 	imes 2^3)^{\frac{3}{5}} \cdot (5^4 	imes 2^2)^{\frac{2}{5}} \cdot (3 	imes 2^2 	imes 5^2)^{\frac{1}{5}}
   $$

2. **Simplify Exponents**:  
   Applying fractional powers to each term:  
   - $$ (3^3)^{\frac{3}{5}} = 3^{\frac{9}{5}}, (2^3)^{\frac{3}{5}} = 2^{\frac{9}{5}} $$  
   - $$ (5^4)^{\frac{2}{5}} = 5^{\frac{8}{5}}, (2^2)^{\frac{2}{5}} = 2^{\frac{4}{5}} $$  
   - $$ (3)^{\frac{1}{5}}, (2^2)^{\frac{1}{5}} = 2^{\frac{2}{5}}, (5^2)^{\frac{1}{5}} = 5^{\frac{2}{5}} $$

Combining all terms:  
   $$
   (3^{\frac{9}{5} + \frac{1}{5}}) \cdot (2^{\frac{9}{5} + \frac{4}{5} + \frac{2}{5}}) \cdot (5^{\frac{8}{5} + \frac{4}{5} + \frac{2}{5}})
   $$

Simplified powers:  
   - For $$3: \frac{9}{5} + \frac{1}{5} = 2$$, so $$3^2$$.  
   - For $$2: \frac{9}{5} + \frac{4}{5} + \frac{2}{5} = 3$$, so $$2^3$$.  
   - For $$5: \frac{8}{5} + \frac{4}{5} + \frac{2}{5} = 2$$, so $$5^2$$.

Final expression:  
   $$
   3^2 \cdot 2^3 \cdot 5^2
   $$

3. **Count Distinct Prime Factors**:  
   The distinct prime factors are $$3, 2, and 5.$$

Thus, the total number of distinct prime factors is **3**.

Answer

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