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) \[(a-b)\left( b-c \right)\left( c-a \right)\]
D) \[\left( a+b \right)\left( b+c \right)\left( c+a \right)\]
Correct Answer: D
Solution :
(d): We have \[\left( {{a}^{2}}-{{b}^{2}} \right)+\left( {{b}^{2}}-{{c}^{2}} \right)-\left( {{c}^{2}}-{{a}^{2}} \right)=0\] \[\therefore {{({{a}^{2}}-{{b}^{2}})}^{3}}+{{({{b}^{2}}-{{c}^{2}})}^{3}}+{{({{c}^{2}}-{{a}^{2}})}^{3}}\] \[=({{a}^{2}}-{{b}^{2}})({{b}^{2}}-{{c}^{2}})({{c}^{2}}-{{a}^{2}})\] =\[3\left( a-b \right)\left( a+b \right)\left( b-c \right)\left( b+c \right)\left( c-a \right)\left( c+a \right)\] Similarly, we have \[\left( a-b \right)+\left( b-c \right)+\left( c-a \right)=0\] \[\Rightarrow {{\left( a-b \right)}^{3}}+{{\left( b-c \right)}^{3}}+{{\left( c-a \right)}^{3}}\] \[=3\left( a-b \right)\left( b-c \right)\left( c-a \right)\] And proceed.
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