If, 4x + y = 10, and 2xy = 8, and 4x > y, then find the

Solution

2xy = 8 So, xy = (8/2) = 4 ......(i) 4x + y = 10 On squaring the above equation, we get, (4x + y)² = 10² Or, 16x² + y² + 2 × 4x × y = 100 Or, 16x² + y² + 8 × 4 = 100 (since, xy = 4) Or, 16x² + y² = 100 - 32 So, 16x² + y² = 68 ....(ii) (4x - y)² = 16x² + y² - 2 × 4x × y = 68 - 8 × 4 = 68 - 32 = 36 {using values from equations (i) and (ii)} Since, 4x > y, so, 4x - y, cannot be negative. So, 4x - y = 6 ....(iii) We know that, a³ - b³ = (a - b) × (a² + b² + ab). So, (4x)³ - (y)³ = 64x³ - y³ = (4x - y) × (16x² + y² + 4x × y) Putting the value of equations (i), (ii) and (iii) in the above equation, we get, = 6 × (68 + 4 × 4) = 6 × (68 + 16) = 6 × 84 = 504 Hence, option b.