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Question

Find (a2b2)3+(b2c2)3+(c2a2)3(ab)3+(bc)3+(ca)3=

A

3(a+b)(b+c)(c+a)

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B

3(a-b)(b-c)(c-a)

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C

(a-b)(b-c)(c-a)

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D

(a+b)(b+c)(c+a)

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Solution

Since, we have if x+y+z=0 then x3+y3+z3=3xyz
Put x=ab,y=bc,z=ca then
x+y+z=ab+bc+ca=0
(ab)3+(bc)3+(ca)3=3(ab)(bc)(ca).....(i)
Similar way, if x=a2b2,y=b2c2,z=c2a2 then
x+y+z=a2b2+b2c2+c2a2=0
Thus, (a2b2)3+(b2c2)3+(c2a2)3=3(a2b2)(b2c2)(c2a2)....(ii)
Now consider given expression =(a2b2)3+(b2c2)3+(c2a2)3(ab)3+(bc)3+(ca)3
Substitute equation (i) and equation (ii) in above expression, we get
=3(a2b2)(b2c2)(c2a2)3(ab)(bc)(ca) [Since, (a2b2)=(a+b)(ab)]
=3(a+b)(ab)(b+c)(bc)(c+a)(ca)3(ab)(bc)(ca)=(a+b)(b+c)(c+a)
Therefore, the value of the given expression is (a+b)(b+c)(c+a)


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