A) \[(a-b)(b-c)(c-a)(a+b+c)\]
B) \[abc(a+b)(b+c)(c+a)\]
C) \[(a-b)(b-c)(c-a)\]
D) none of the above
Correct Answer: A
Solution :
Let \[\Delta =\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ {{a}^{3}} & {{b}^{3}} & {{c}^{3}} \\ \end{matrix} \right|\] Applying \[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}},{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}\] \[=\left| \begin{matrix} 1 & 0 & 0 \\ {{a}^{3}} & b-a & c-a \\ {{a}^{3}} & {{b}^{3}}-{{a}^{3}} & {{c}^{3}}-{{a}^{3}} \\ \end{matrix} \right|\] \[=(b-a)(c-a)\] \[\times \,\,\left| \begin{matrix} 1 & 0 & 0 \\ a & 1 & 1 \\ {{a}^{3}} & {{b}^{2}}+ab+{{a}^{2}} & {{c}^{2}}+ac+{{a}^{2}} \\ \end{matrix} \right|\] \[=(b-a)(c-a)\] \[\times \,\,[({{c}^{2}}+ac+{{a}^{2}})-({{b}^{2}}+ab+{{a}^{2}})]\] \[=(b-a)(c-a)[{{c}^{2}}-{{b}^{2}}+ac-ab]\] \[=(b-a)(c-a)(c-b)[c+b+a]\] Note: The value of the determinant does not change either by any rows or by any columns operation.
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