Question

Using properties of determinant prove that |[1, a, a^2],[1, b, b^2],[1, c, c^2]| = (a-b)(b-c)(c-a)

Accepted Answer

The determinant of a matrix  [[a, b, c],[d,e,f],[g,h,i]] is given by:
|[a, b, c],[d,e,f],[g,h,i]|= a*(e*i - f*g) - b*(d*i - g*f) + c*(d*h -e*g) 
The determinant that has to be found |[1, a, a^2], [1, b, b^2],[1, c, c^2]|
= 1*(b*c^2 - b^2*c) - a*(c^2 - b^2) + a^2*(c - b) 
=> b*c^2 - b^2*c + a*b^2 - a*c^2 + a^2*c - a^2*b
=> bc(c - b) + a(b - c)(b + c) + a^2(c - b) 
=> (b - c)(ab + ac - bc - a^2) 
=> (b - c)(a(c - a) -b(c - a)) 
=> (a - b)(b - c)(c - a) 
This proves that |[1, a, a^2], [1, b, b^2],[1, c, c^2]| = (a - b)(b - c)(c - a)

Answer

1 a a^2
1 b b^2
1 c c^2      appyling the R1= (R1-R2) and R2=(R2-R3)
then,    0  (a-b)  (a^2-b^2)
           0  (b-c)  (b^2-c^2)
           1    c         c^2
taking{ (a-b) (b-c) }common from 
(a-b)(b-c) , 0 1 (a+b)
                 0 1 (b+c)
                 1  c  c^2
  expanding by C1
            (a-b)(b-c)(a-c)

Math

Using properties of determinant prove that `|[1, a, a^2],[1, b, b^2],[1, c, c^2]| = (a-b)(b-c)(c-a)`

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