# How to solve; Factorise a^2(b+c) + b^2(c+a)+c^2(a+b)+2abc ?

Welcome to my article How to solve; Factorise a^2(b+c) + b^2(c+a)+c^2(a+b)+2abc ? This question is taken from the simplification lesson.
The solution of this question has been explained in a very simple way by a well-known teacher by doing addition, subtraction, and fractions.
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## How to solve; Factorise a^2(b+c) + b^2(c+a)+c^2(a+b)+2abc ?

When we write it like this,

\displaystyle {{a}^{2}}(b+c)+{{b}^{2}}(c+a)+{{c}^{2}}(a+b)+2abc

\displaystyle {{a}^{2}}b+{{a}^{2}}c+{{b}^{2}}c+{{b}^{2}}a+{{c}^{2}}a+{{c}^{2}}b+2abc

\displaystyle {{b}^{2}}a+{{c}^{2}}a+2abc+{{a}^{2}}b+{{a}^{2}}c+{{b}^{2}}c+{{c}^{2}}b

\displaystyle a({{b}^{2}}+{{c}^{2}}+2bc)+{{a}^{2}}b+{{a}^{2}}c+{{b}^{2}}c+{{c}^{2}}b

we know that ,

\displaystyle {{(A+B)}^{2}}={{A}^{2}}+{{B}^{2}}+2AB

so,

\displaystyle a{{(b+c)}^{2}}+{{a}^{2}}b+{{a}^{2}}c+{{b}^{2}}c+{{c}^{2}}b

\displaystyle a{{(b+c)}^{2}}+{{a}^{2}}(b+c)+bc(b+c)

\displaystyle a(b+c)(b+c)+{{a}^{2}}(b+c)+bc(b+c)

\displaystyle (b+c)[a(b+c)+{{a}^{2}}+bc]

\displaystyle (b+c)[ab+ac+{{a}^{2}}+bc]

\displaystyle (b+c)[ab+bc+ac+{{a}^{2}}]

\displaystyle (b+c)[b(a+c)+a(c+a)]

\displaystyle (b+c)[b(a+c)+a(a+c)]