Question

solution Let the value of the given determinant be ( Delta ). Then,
[
begin{aligned}
Delta &=left|begin{array}{ccc}
1 & 1 & 1
a & b & c
a^{3} & b^{3} & c^{3}
end{array}right|
&=left|begin{array}{ccc}
0 & 0 & 1
a-c & b-c & c
a^{3}-c^{3} & b^{3}-c^{3} & c^{3}
end{array}right|
end{aligned}
]
[applying ( left.C_{1} rightarrowleft(C_{1}-C_{3}right) text { and } C_{2} rightarrowleft(C_{2}-C_{3}right)right] )
[
=(a-c)(b-c) cdotleft|begin{array}{ccc}
0 & 0 & 1
1 & 1 & c
a^{2}+a c+c^{2} & b^{2}+b c+c^{2} & c^{3}
end{array}right|
]
( left[text { taking out }(a-c) text { and }(b-c) text { common from } C_{1} text { and } C_{2}right] ) ( left.=(a-c)(b-c) cdot 1 cdotleft|begin{array}{cc}1 & 1 a^{2}+a c+c^{2} & b^{2}+b c+c^{2}end{array}right| text { [expanded by } R_{1}right] )
( =(a-c)(b-c) cdotleft[left(b^{2}+b c+c^{2}right)-left(a^{2}+a c+c^{2}right)right] )
( =(a-c)(b-c)left[left(b^{2}-a^{2}right)+(b c-a c)right] )
( =(a-c)(b-c)left[left(b^{2}-a^{2}right)+(b-a) cright] )
( =(a-c)(b-c)(b-a)(b+a+c) )
[
=(a-b)(b-c)(c-a)(a+b+c)
]
Hence, ( Delta=(a-b)(b-c)(c-a)(a+b+c) )

# 1 1 a 1 b 1 1 c = (a - b)(b-c)(c-a)(a + b + c). EXAMPLES Prove that (CBSE 2009C

Solution