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  • prove that: (a+b+c)^{3}-a^{3}-b^{3}-c^{3}=3(a+b)(b+c)(c+a)

prove that: (a+b+c)^{3}-a^{3}-b^{3}-c^{3}=3(a+b)(b+c)(c+a)

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prove that: (a+b+c)^{3}-a^{3}-b^{3}-c^{3}=3(a+b)(b+c)(c+a)

Taking left hand side

(a+b+c)^{3}-a^{3}-b^{3}-c^{3}=\left ( (a+b+c)^{3}-a^{3} \right )\left ( b^{3}+c^{3} \right )\cdots \cdots \cdots \cdots (i)              

Now using the identity

x^{3}-y^{3}=(x-y)(x^{2}+y^{2}+xy)

So,

\left ( (a+b+c)^{3}-a^{3} \right )=(a+b+c-a)\left [ (a+b+c)^{2}+a^{2}+a(a+b+c) \right ]

Now using x^{3}-y^{3}=(x-y)(x^{2}+y^{2}+xy)

 (b^{3}+c^{3})=\left [ (b+c)(b^{2}+c^{2}-bc) \right ]

 So equation (i) becomes:

\\(a+b+c)^{3}-a^{3}-b^{3}-c^{3}\\ =(a+b+c-a)\left [ (a+b+c)^{2}+a^{2}+a(a+b+c) \right ]-\left [ (b+c)(b^{2}+c^{2}-bc) \right ]\\ =(b+c)\left [ (a+b+c)^{2}+a^{2}+a(a+b+c)\right ]\left [ (b+c)(b^{2}+c^{2}-bc) \right ]\cdots \cdots \cdots (ii)\\

Now,

(a+b+c)^{3}-a^{3}-b^{3}-c^{3}\\ =(b+c)\left [ a^{2}+b^{2}+c^{2}+2ab+2bc+2ca+a^{2}+a^{2}+ab+ac-(b^{2}+c^{2}-bc) \right ]\\ =(b+c)\left [ b^{2}+c^{2}+3a^{2}+3ab+3ac-b^{2}-c^{2}+3bc \right ]\\ =(b+c)\left [ 3(a^{2})+ab+ac+bc \right ]\\ =3(b+c)\left [ a(a+b)+c(a+b) \right ]\\ =3(b+c)\left [ (a+b)(b+c) \right ]\\ =3(a+b)(b+c)(c+a)=R.H.S

Hence proved.

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