evaluate 7(a square-b square) cube+7(b square-c square) cube +7( c square-a square) cube by (a-b) cube+(b-c) cube+(c-a) cube
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Solution:
/* we know that,
if x+y+z = 0 then x³+y³+z³ = 3xyz */
i) [(a²-b²)³+(b²-c²)³+(c²-a²)]/[(a-b)³+(b-c)³+(c-a)³]
= [3(a²-b²)(b²-c²)(c²-a²)]/[3(a-b)(b-c)(c-a)]
=[(a+b)(a-b)(b+c)(b-c)(c+a)(c-a)]/[(a-b)(b-c)(c-a)]
After cancellation,we get
= (a+b)(b+c)(c+a)
••••
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Answer:
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