Question

SOLUTION The given determinant
( =left|begin{array}{lll}1 & b c & a(b+c) 1 & c a & b(c+a) 1 & a b & c(a+b)end{array}right| )
( left.=left|begin{array}{ccc}1 & b c & a b+b c+c a 1 & c a & a b+b c+c a 1 & a b & a b+b c+c aend{array}right| text { [applying } C_{3} rightarrow C_{3}+C_{2}right] )
( left.=(a b+b c+c a) cdotleft|begin{array}{ccc}1 & b c & 1 1 & c a & 1 1 & a b & 1end{array}right| text { [taking }(a b+b c+c a) text { common from } C_{3}right] )
( =(a b+b c+c a) times 0=0 )
( left[because C_{1} text { and } C_{3} ) are identical] right.

# 11 bca(b + c) EXAMPLE 4 Prove that 1 ca b(c+a) = 0. 11 ab c(a+b)|

Solution