If a+b+c=0, then the value of (b+c)2bc+(c+a)2ca+(a+b)2ab
A
0
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B
1
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C
2
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D
3
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Solution
The correct option is D
3 Given a+b+c=0⇒b+c=−a,c+a=−b,a+b=−c(b+c)2bc+(c+a)2ca+(a+b)2ab=(−a)2bc+(−b)2ca+(−c)2ab=a2bc+(−b)2ca+(−c)2ab=a2bc+b2ca+c2ab=a3+b3+c3abc=3abcabc=3
Hence, the correct answer is option (d)