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Well done. A common appraoch students make is to first determine the values of x , y , z which is not always the case for all system of equations.
we have:A= (x+1)(y+1)(z+1)=xyz+x+y+z+xy+yz+zx+1, so when we plus both 3 equations, we have 3(xyz+x+y+z+xy+yz+zx)=267 <=> 3(A-1)=267, so A=90
Superb!
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Expanding ( x + 1 ) ( y + 1 ) ( z + 1 ) = x y z + x y + y z + z x + x + y + z + 1
We note that adding the three given equations we have:
⎩ ⎪ ⎨ ⎪ ⎧ x y z + 3 x + 3 y z = 1 0 6 x y z + 3 z + 3 x y = 7 9 x y z + 3 y + 3 z x = 8 2 ⇒ 3 ( x y z + x y + y z + z x + x + y + z ) = 2 6 7
⇒ ( x + 1 ) ( y + 1 ) ( z + 1 ) = 3 2 6 7 + 1 = 8 9 + 1 = 9 0