A) \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc\]
B) \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}+3abc\]
C) \[(a+b+c)(a-b)(b-c)(c-a)\]
D) None of these
Correct Answer: C
Solution :
\[\Delta =\left| \,\begin{matrix} 1 & 1 & 1 \\ a & b & c \\ {{a}^{3}} & {{b}^{3}} & {{c}^{3}} \\ \end{matrix}\, \right|\] vanishes when \[a=b,\,b=c,\,c=a\]. Hence \[(a-b),\,(b-c),\,(c-a)\] are factors of \[\Delta \]. Since \[\Delta \] is symmetric in \[{{C}_{21}}={{(-1)}^{2+1}}(18+21)=-39\] and of 4th degree, \[(a+b+c)\] is also a factor, so that we can write \[-10x+90-42-81+42+9x=0\,\,\text{or}\,\text{ }x=9\] ......(i) Where by comparing the coefficients of the leading term \[b{{c}^{3}}\] on both the sides of identity (i). We get \[1=k(-1)\,(-1)\Rightarrow k=1\] \[{{C}_{3}}\to {{C}_{3}}-{{C}_{1}},\]. Trick: Put\[a=1,\,b=2,\,c=3\], so that determinant \[\left| \,\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 8 & 27 \\ \end{matrix}\, \right|=1(30)-1(24)+1(8-2)=12\] which is given by (c). i.e. \[(1+2+3)\,(1-2)\,(2-3)(3-1)=12\].
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