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Question

Factorize :  $$a^{2}(b+c)^{2}+b^{2}(c+a)^{2}+c^{2}(a+b)+abc(a+b+c)+(a^{2}+b^{2}+c^{2})(bc+ca+ab)$$


Solution

$$f(a)=a^{2}(b+c)^{2}+b^{2}(c+a)^{2}+c^{2}(a+b)^{2}+abc(a+b+c)+(a^{2}+b^{2}+c^{2})(bc+ca+ab)$$
$$f(-b)=b^{2}(b+c)^{2}+b^{2}(c-b)^{2}-b^{2}c^{2}+(2b^{2}+c^{2})(bc-bc-b^{2})$$
$$=b^{2}(b^{2}+c^{2}+2b^{4}c^{2}+b^{2}-2bc)$$
$$=b^{2}\left [ 2b^{2}+2c^{2} \right ]$$
$$=2b^{4}+| 2b^{2}c^{2}-b^{2}c^{2}-2b^{4}-b^{2}c^{2}=0$$
$$\therefore (a+b)$$ is a factor
Since the expression is cyclic,
$$(a+b)(b+c)(c+a)$$ are factors.
$$\therefore k(a+b+c)(a+b)(b+c)(c+a)$$
By putting $$a=0, b=1, c=2,$$ we get
$$k=1$$
$$\therefore (a+b+c)(a+b)(b+c)(c+a)$$ are the factors.

Mathematics

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