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Q.

3a+ b+ ca2 + b2 + c2a + b+ ca2 + b2 + c2a3 + b3 + c3a2 + b2 + c2a3 + b3 + c3a4 + b4 + c4 = k(a-b)2(b-c)2(c-a)2then k =

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a

4

b

3

c

1

d

0

answer is C.

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Detailed Solution

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Let a=0, b=1, c=1

302020202=k(0-1)2 (2)2 (-1)2        12-8=4k             k=1

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3a+ b+ ca2 + b2 + c2a + b+ ca2 + b2 + c2a3 + b3 + c3a2 + b2 + c2a3 + b3 + c3a4 + b4 + c4 = k(a-b)2(b-c)2(c-a)2then k =