(a) x + y + z + 2 = xyz.
x = logabc ⇒ ax = bc ⇒ ax + 1 = abc ⇒ a = (abc)1/(x +1)
Similarly, b = (abc)1/(y + 1), c = (abc)1/(z + 1)
∴ abc = \((abc)^{\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}}\)
⇒ \({\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}}\) = 1
⇒ (y + 1) (z + 1) + (x + 1) (z + 1) + (x + 1) (y + 1) = (x + 1) (y + 1) (z + 1)
⇒ yz + y + z + 1 + xz + x + z + 1 + xy + y + x + 1 = xyz + xy + yz + zx + x + y + z + 1
⇒ x + y + z + 2 = xyz.