Question

Using properties of determinant prove that 

`|(b+c , a , a), (b , c+a, b), (c, c, a+b)|` = 4abc

Answer 1

Let Δ = `|(b+c , a , a), (b , c+a, b), (c, c, a+b)|`

R₁ → R₁ - R₂ - R₃  

Δ = `|(0 ,-2c ,-2b), (b , c+a, b), (c, c, a+b)|`

Expending R₁

Δ=`0|(c+a, b),(c, a+b)| -(-2c)|(b, b),(c, a+b)| +(-2b)|(b, c+a),(c, c)|`

= 2c(ab + b² - bc) - 2b(bc - c² - ac)

= 2abc + 2cb² - 2bc² - 2b²c + 2bc² + 2abc

= 4abc