Question

Prove that:

`sqrt(3xx5^-3)divroot3(3^-1)sqrt5xxroot6(3xx5^6)=3/5`

Answer 1

we have to prove that `sqrt(3xx5^-3)/(root3(3^-1)sqrt5)xxroot6(3xx5^6)=3/5`

By using rational exponents `a^-n=1/a^n` we get,

`sqrt(3xx5^-3)/(root3(3^-1)sqrt5)xxroot6(3xx5^6)=sqrt(3xx1/5^3)/(root3(1/3)sqrt5)xxroot6(3xx5^6)`

`=(3^(1/2)xx1/5^(3xx1/2))/(1/3^(1/3)xx5^(1/2))xx3^(1/6)xx5^(6xx1/6)`

`=(3^(1/2)/5^(3/2))/(5^(1/2)/3^(1/3))xx3^(1/6)xx5^1`

`=3^(1/2)/5^(3/2)xx3^(1/3)/5^(1/2)xx3^(1/6)xx5^1`

`=3^(1/2)xx3^(1/3)xx5^(-3/2)xx5^(-1/2)xx3^(1/6)xx5^1`

`=3^(1/2+1/3+1/6)xx5^(-3/2-1/2+1)`

`=3^((1xx3)/(2xx3)+(1xx2)/(3xx2)+1/6)xx5^(-3/2-1/2+(1xx2)/(1xx2))`

`=3^((3+2+1)/6)xx5^((-3-1+2)/2)`

`=3^1xx5^-1`

`=3xx1/5`

`=3/5`

Hence `sqrt(3xx5^-3)/(root3(3^-1)sqrt5)xxroot6(3xx5^6)=3/5`