Question

Number of distinct prime factors in
(729)^(2/5) X (4125)^(3/5) X (605)^(1/5) is ____​

Answer 1

Answer:

3 , 5 & 11

Step-by-step explanation:

Number of distinct prime factors in

(729)^(2/5) X (4125)^(3/5) X (605)^(1/5) is _

729 = 3^6

729^2 = 3^12

4125 = 3 × 5^3 × 11

4125^3 = 3^3 × 5^9 × 11^3

605 = 5 × 11^2

(729)^(2/5) X (4125)^(3/5) X (605)^(1/5)

= (3^12)^(1/5)× (3^3 × 5^9 × 11^3)^(1/5) × (5 × 11^2)^(1/5)

= (3^15 × 5^10 × 11^5)^(1/5)

= (3 ^3 × 5^2 × 11)

Distinct prime factor are 3 , 5 & 11