If a and b are different positive primes such that$ (a+b)^{-1}\left(a^{-1}+b^{-1}\right)=a^{x} b^{y} $, find $ x+y+2 . $

Given:

a and b are different positive primes such that

\( (a+b)^{-1}\left(a^{-1}+b^{-1}\right)=a^{x} b^{y} \).

To do: 

We have to find \( x+y+2 . \)

Solution:

We know that,

$(a^{m})^{n}=a^{m n}$

$a^{m} \times a^{n}=a^{m+n}$

$a^{m} \div a^{n}=a^{m-n}$

$a^{0}=1$

Therefore,

$(a+b)^{-1} (a^{-1}+b^{-1})=a^{x} b^{y}$

$(a+b)^{-1} (\frac{1}{a}+\frac{1}{b})=a^{x} b^{y}$

$\frac{1}{a+b} \times \frac{b+a}{a b}=a^{x} b^{y}$

$\frac{1}{a b}=a^{x} b^{y}$

$a^{-1} b^{-1}=a^{x} b^{y}$

Comparing both sides, we get,

$x=-1$ and $y=-1$

Therefore,

$x+y+2=-1-1+2$

$=0$

The value of $x+y+2$ is $0$.