A) \[3\left( a+b \right)\left( b+c \right)\left( c+a \right)\]
B) \[3\left( a-b \right)\left( b-c \right)\left( c-a \right)\]
C) \[\left( a-b \right)\left( b-c \right)\left( c-a \right)\]
D) \[\left( a+b \right)\left( b+c \right)\left( c+a \right)\]
Correct Answer: D
Solution :
(d): Expanding numerator (N) \[\cancel{{{a}^{6}}}-\cancel{{{b}^{6}}}-3{{a}^{2}}{{b}^{2}}({{a}^{2}}-{{b}^{2}})\] \[+\cancel{{{b}^{6}}}-\cancel{{{c}^{6}}}-3{{b}^{2}}{{c}^{2}}({{b}^{2}}-{{c}^{2}})\] \[+\cancel{{{c}^{6}}}-\cancel{{{a}^{6}}}-3{{c}^{2}}{{a}^{2}}({{c}^{2}}-{{a}^{2}})\] \[=-\,3\left[ \begin{align} & ab\left( a+b \right)\times ab\left( a-b \right)+bc\left( b+c \right)\times bc\left( b-c \right) \\ & +ca\left( c+a \right)\times ca\left( c-a \right) \\ \end{align} \right]\]Expanding denominator, (d) \[\cancel{{{a}^{3}}}-\cancel{{{b}^{3}}}-3ab(a-b)\] \[+\cancel{{{b}^{3}}}-\cancel{{{c}^{3}}}-3bc(b-c)\] \[+\cancel{{{c}^{3}}}-\cancel{{{a}^{3}}}-3ca(c-a)\] \[=-3\left[ ab\left( a-b \right)+bc\left( b-c \right)+ca\left( c-a \right) \right]\] Now, dividing N/D, you get \[\left( a+b \right)\left( b+c \right)(c+a)\]
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