If a,b,c are all distinct and `|[a,a^3,a^4-1],[b,b^3,b^4-1],[c,c^3,c^4-1]|` =0, show that abc(ab+bc+ac) = a+b+c

Explanation

Given,`|[a,a^3,a^4-1],[b,b^3,b^4-1],[c,c^3,c^4-1]|=0`

Doing Row operation `R_1​->R_1​−R_2​,R_3​->R_3​−R_2`

​=`|[a-b,a^3-b^3,a^4-b^4],[b,b^3,b^4-1],[c-b,c^3-b^3,c^4-b^4]| =0`

​=`|[a-b,(a−b)(a^2+b^2+ab),(a^2+b^2)(a+b)(a−b)],[b,b^3,b^4-1],[c-b,(c−b)(c^2+b^2+cb),(c^2+b^2)(c+b)(c−b)]| =0`

Taking out (a−b) and (c−b) from the determinant and Doing Row operations` R_1​->R_1​−R_3`

​ ​=`(a-b)(c-b)|[0,a^2-c^2−b(a-c),(a^3−c^3)+b^2(a−c)+b(a^2−c2)],[b,b^3,b^4-1],[1,(c^2+b^2+cb),(c^2+b^2)(c+b)]| =0`

​=`(a-b)(c-b)(c-a)|[0,a+b+c,(a^2+c^2+ac)+b^2+b(a+c)],[b,b^3,b^4-1],[1,(c^2+b^2+cb),(c^2+b^2)(c+b)]| =0`

Since a,b,c are distinct.

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