Prove the following : ∣∣
∣
∣∣b2+c2abacabc2+a2bccacba2+b2∣∣
∣
∣∣ =∣∣
∣
∣∣b2+c2a2a2b2c2+a2b2c2c2a2+b2∣∣
∣
∣∣=4a2b2c2
Open in App
Solution
Take out a, b, c from C1, C2 and C3 respectively. △=abc∣∣
∣∣aca+ca+bbabb+cc∣∣
∣∣ Apply C1+(C2−C3) and take 2b common from C1 △=abc.2b∣∣
∣∣0ca+c1ba1b+cc∣∣
∣∣ Apply R2−R3 △=2ab2c∣∣
∣∣0ca+c0−ca−c1b+cc∣∣
∣∣ =2ab2c{c(a−c)+c(a+c)} =2ab2c(2ac)=4a2b2c2