NDA Mathematics Question (2025) — Solution

Step-by-step solution

Given x = a + b, y = a + b ^2, and z = a ^2 + b . Squaring each term, we get: x^2 = (a + b)^2 = a^2 + b^2 + 2ab y^2 = (a + b ^2)^2 = a^2 ^2 + b^2 ^4 + 2ab ^3 z^2 = (a ^2 + b )^2 = a^2 ^4 + b^2 ^2 + 2ab ^3 Using the properties of cube roots of unity, ^3 = 1 and ^4 = . Substituting these into the equations: y^2 = a^2 ^2 + b^2 + 2ab z^2 = a^2 + b^2 ^2 + 2ab Adding x^2, y^2, and z^2: x^2 + y^2 + z^2 = a^2(1 + + ^2) + b^2(1 + + ^2) + 2ab(1 + 1 + 1) Since 1 + + ^2 = 0, the expression simplifies to: x^2 + y^2 + z^2 = a^2(0) + b^2(0) + 6ab = 6ab