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Factorise : (a+b)^3+(b+c)^3+(c+a)^3-3(a+...

Factorise : `(a+b)^3+(b+c)^3+(c+a)^3-3(a+b)(b+c)(c+a)`

Text Solution

Verified by Experts

We know that
`a^3+b^3+c^3−3abc=(a+b+c)(a^2+b^2+c^2−ab−bc−ca)`
i.e. ,
`=>``(a+b)^3+(b+c)^3+(c+a)^3−3(a+b)(b+c)(c+a)=((a+b)+(b+c)+(c+a))[(a+b)^2+(b+c)^2+(c+a)^2−(a+b)(b+c)−(b+c)(c+a)−(c+a)(a+b)]`
`=>``(a+b)^3+(b+c)^3+(c+a)^3-3(a+b)(b+c)(c+a)=2(a+b+c)[a^2+b^2+c^2-ab-bc-ca]`
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