Question

sourion Let the given determinant be ( Delta ) Then,
[
begin{array}{l}
Delta=left|begin{array}{ccc}
1 & a & a^{2}
1 & b & b^{2}
1 & c & c^{2}
end{array}right|
left.begin{array}{rl}
1 & a & a^{2}
0 & b-a & b^{2}-a^{2}
0 & c-a & c^{2}-a^{2}
end{array} mid text { [applying } R_{2} rightarrowleft(R_{2}-R_{1}right) text { and } R_{3} rightarrowleft(R_{3}-R_{1}right)right]
end{array}
]
( =(b-a)(c-a) cdotleft|begin{array}{ccc}1 & a & a^{2} 0 & 1 & b+a 0 & 1 & c+aend{array}right| )
( left[operatorname{taking}(b-a) text { common from } R_{2} text { and }(c-a) text { common from } R_{3}right] ) ( left.=(b-a)(c-a) times 1 cdotleft|begin{array}{cc}1 & b+a 1 & c+aend{array}right| text { [expanded by } C_{1}right] )
( =(b-a)(c-a){(c+a)-(b+a)} )
( =(b-a)(c-a)(c-b)=(a-b)(b-c)(c-a) )
Hence, ( Delta=(a-b)(b-c)(c-a) )

# EXAMPLE 7 Prove that 1 b_b2 = (a - b)(b -c)(c-a). 1 ca

Solution