The number of positive integral solutions of the equation x3+1x2yx2zxy2y3+1y2zxz2yz2z3+1=11 is

x3+1x2yx2zxy2y3+1y2zxz2yz2z3+1=11Multiplying R1 by x, R2 by y and R3 by z, we get1xyzx4+xx3yx3zxy3y4+yy3zxz3yz3z4+z=11Taking x, y, z common from C1, C2, C3, respectively, we getx3+1x3x3y3y3+1y3z3z3z3+1=11Using R1→R1+R2+R3, we havex3+y3+z3+1111y3y3+1y3z3z3z3+1=11Using C2→C2−C1 and C3→C3−C1, we getx3+y3+z3+1100y310z301=11Hence, x3 + y3 + z3 = 10Therefore, the ordered triplets are (2, 1, 1), (1, 2, 1), (1, 1, 2).