Question

(a-3b)^3+(3b-2c)^3+(2c-a)^3​

Answer 1

(a-3b)^3+(3b-2c)^3+(2c-a)^3

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Answer 2

Answer:

We know the corollary: if a+b+c=0 then a^3 + b^3 + c^3 = 3abc

Using the above corollary taking a=(a−3b), b=(3b−c) and c=(c−a), we have a+b+c=a−3b+3b−c+c−a=0 then the value of (a−3b)^3+(3b−c)^3+(c−a)^3 is:

(a−3b)^3+(3b−c)^3+(c−a)^3

=3[(a−3b)×(3b−c)×(c−a)]=3(a−3b)(3b−c)(c−a)

Hence, (a−3b)^3+(3b−c)^3+(c−a)^3

=3(a−3b)(3b−c)(c−a)

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