```
ll a=2,b=3,x=5;
ll n=a^x+b^x;
cout<<n<<" ";
ll m=a^x;n=b^x;
cout<<n+m;
```

why it outputs different output both time even I’am doing same thing.

someone plzz help

```
ll a=2,b=3,x=5;
ll n=a^x+b^x;
cout<<n<<" ";
ll m=a^x;n=b^x;
cout<<n+m;
```

why it outputs different output both time even I’am doing same thing.

someone plzz help

Operator precedence.

due to precedence is expression is evaluated as : (a)^(x+b)^(b). so the ans is 15:

use. this ` ll n=(a^x)+(b^x);`

ohh i got it now thanks all of you