Algebra (Olympiad)hardFree

Solve (x-1)(y-1)(z-1)=xyz-1 with the (x-2) System | IOQM

JEE Maths question with a full step-by-step solution.

Question
Solve the system in real numbers:
(x−1)(y−1)(z−1)=xyz−1and(x−2)(y−2)(z−2)=xyz−2.(x - 1)(y - 1)(z - 1) = xyz - 1 \quad \text{and} \quad (x - 2)(y - 2)(z - 2) = xyz - 2.
A(x,y,z)=(1,0,1)(x, y, z) = (1, 0, 1).
B(x,y,z)=(1,2,1)(x, y, z) = (1, 2, 1).
C(x,y,z)=−1,1,1)(x, y, z) = -1, 1, 1).
D(x,y,z)=(1,1,1)(x, y, z) = (1, 1, 1).correct
Solution
Step 1: Expand the first equation.
xyz−(xy+yz+zx)+(x+y+z)−1=xyz−1  ⟹  xy+yz+zx=x+y+z.xyz - (xy + yz + zx) + (x + y + z) - 1 = xyz - 1 \implies xy + yz + zx = x + y + z.
 Let this common value kk. ...(1) Step 2: Expand the second equation.
xyz−2(xy+yz+zx)+4(x+y+z)−8=xyz−2xyz - 2(xy + yz + zx) + 4(x + y + z) - 8 = xyz - 2
  ⟹  −2(xy+yz+zx)+4(x+y+z)=6.\implies -2(xy + yz + zx) + 4(x + y + z) = 6.
 Using (1), −2k+4k=6-2k + 4k = 6, so k=3k = 3. Thus xy+yz+zx=x+y+z=3xy + yz + zx = x + y + z = 3. Step 3: Compute x2+y2+z2x^2 + y^2 + z^2.
x2+y2+z2=(x+y+z)2−2(xy+yz+zx)=32−2×3=3.x^2 + y^2 + z^2 = (x + y + z)^2 - 2(xy + yz + zx) = 3^2 - 2 \times 3 = 3.
 Step 4: Since x2+y2+z2=3=xy+yz+zxx^2 + y^2 + z^2 = 3 = xy + yz + zx,
12[(x−y)2+(y−z)2+(z−x)2]=(x2+y2+z2)−(xy+yz+zx)=0.\frac{1}{2}\big[(x - y)^2 + (y - z)^2 + (z - x)^2\big] = (x^2 + y^2 + z^2) - (xy + yz + zx) = 0.
 This forces x=y=zx = y = z. With x+y+z=3x + y + z = 3, we get x=y=z=1x = y = z = 1. Answer: (x,y,z)=(1,1,1)(x, y, z) = (1, 1, 1).
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