xyz+xy+xz+yz+x+y+z=384,find x+y+z where x,y,z are positive integers....plz answer quickly
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Xyz+xy+xz+yz+x+y+z=384
xy (z+1) + z (x+y)+1(x+y)+z=384
xy (z+1) + (x+y) (z+1) +(z+1)−1=384
xy (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)=5×7×11
On 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 case
x+1+y+1+z+1=5+11+7x+y+z+3=23
x+y+z=20 ANS
HOPE IT HELPS
xy (z+1) + z (x+y)+1(x+y)+z=384
xy (z+1) + (x+y) (z+1) +(z+1)−1=384
xy (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)=5×7×11
On 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 case
x+1+y+1+z+1=5+11+7x+y+z+3=23
x+y+z=20 ANS
HOPE IT HELPS
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