If `((x^(-1)y^2)/(x^3y^(-2)))^1.\ ((x^6\ y^(-3))/(x^2\ y^3))^(1/2)=x^a y^b ,` prove that `a+b=-1` , where `x\ a n d\ y` are different positive primes.

Explanation

`((x^-1y^2)/(x^3y^(-2)))^1((x^6y^- 3)/(x^2y^3))^(1/2)=x^ay^b`
`rArr(x^(-1-3)y^(2+2))^1(x^(6-2)y^(-3-3))^ (1/2)=x^ay^b`
`rArr(x^-4y^4)^1(x^4y^-6)^1/2=x^ay^b`
`rArr(x^-4y^4)^1(x^2y^-3)^(2xx1/2)=x^ay^b`
`rArr(x^-4y^4)^1(x^2y^-3)^1=x^ay^b``rArrx^ay^b=x^-2y^1`
Comparing both sides, we get
`a=-2` and `b=1`
`thereforea+b=-2+1=-1`
Hence Proved.