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`        Xyz+xy+yz+zx+x+y+z =384 find x+y+zFind answer................?..................?....................?  ..        .......              ............     ......`
2 years ago

Arun
23312 Points
```							Dear Sneha Xyz+xy+xz+yz+x+y+z=384xy (z+1) + z (x+y)+1(x+y)+z=384xy (z+1) + (x+y) (z+1) +(z+1)−1=384xy (z+1) + (x+y) (z+1) +(z+1)=385(z+1) (xy+x+y+1)=385(z+1) (x(y+1)+1(y+1))=385(x+1)(y+1)(z+1)=385(x+1)(y+1)(z+1)=5×7×11On comparing ,x+1 = 5 , y+1=7 , z+1=11or x+1 = 7 , y+1=11 , z+1=5or x+1 = 11 , y+1=5 , z+1=7Now , For each casex+1+y+1+z+1=5+11+7x+y+z+3=23x+y+z=20   ANS  RegardsArun (askIITians forum expert)
```
2 years ago
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### Course Features

• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions