Applying R_{1} → R_{1} + R_{2} + R_{3}, we have

Taking out 3(a + b) from 1st row, we have

Applying C_{1} → C_{1} – C_{2} and C_{2} → C_{2} – C_{3}

Expanding along first row, we have

D = 3(a + b) [1. (4b^{2} – b^{2})]

= 3(a + b) x 3b^{2} = 9b^{2}(a + b)