If, 6x + y = 20, and 2xy = 32, and 6x > y, then find

Solution

2xy = 32 So, xy = (32/2) = 16 ......(i) 6x + y = 20 On squaring the above equation, we get, (6x + y)² = 20² Or, 36x² + y² + 2 × 6x × y = 400 Or, 36x² + y² + 12 × 16 = 400 (since, xy = 16) Or, 36x² + y² = 400 - 192 So, 36x² + y² = 208 ....(ii) (6x - y)² = 36x² + y² - 2 × 6x × y = 208 - 12 × 16 = 208 - 192 = 16 {using values from equations (i) and (ii)} Since, 6x > y, so, 6x - y, cannot be negative. So, 6x - y = 4 ....(iii) We know that, a³ - b³ = (a - b) × (a² + b² + ab). So, (6x)³ - (y)³ = 216x³ - y³ = (6x - y) × (36x² + y² + 6x × y) Putting the value of equations (i), (ii) and (iii) in the above equation, we get, = 4 × (208 + 6 × 16) = 4 × (208 + 96) = 4 × 304 = 1,216 Hence, option b.