Question

Using properties of determinants prove that: |(b + c,a,a),(b,c + a,b),(c,c,a + b)| = 4abc

Answer 1

Considering LHS

On further calculation

= 0 + 2c [b(a + b) – 2c] – 2b [cb – c(c + a)]

By multiplication

= 2c (ab + b² – bc) – 2b (bc – c² – ca)

We get

= 2 abc + 2b²c – 2bc² – 2b²c + 2bc² + 2abc

= 4abc

= RHS

Hence, it is proved.