Question

(a+b+c) (b+c+a) (c+a-b) (a+b-c)​

Answer 1

Step-by-step explanation:

(a+b+c)(b+c+a)=ab+ac+bc+ab+bc+ac+a²+b²+c²

(c+a-b)(a+b-c)=ac+ab+ab-ac-ab+bc+a²-b²-c2

=a²-b²-c²+ab+bc

Final answer=[(ab+ac+bc+ab+bc+ac+a²+b²+c²)(a²-b²-c²+ab+bc)]

Answer 2

Step-by-step explanation:

☆Hey friend!!!☆

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Here is your answer ☞

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$\begin{gathered}\sf \frac{(b + c)}{(a - b)(a - c)} + \frac{(c + a)}{(b - a)(b - c)} + \frac{(a + b)}{(c - a)(c - b)} \\ \\ = \frac{(b + c)}{(a - b)(a - c)} - \frac{(c + a)}{(a - b)(b - c)} + \frac{(a + b)}{(a - c)(b - c)} \\ \\ = \frac{(b + c)(b - c) - (a + c)(a - c) + (a + b)(a - b)}{(a - b)(b - c)(a - c)} \\ \\\sf = \frac{( {b}^{2} - {c}^{2}) - ( {a}^{2} - {c}^{2} ) + ( {a}^{2} - {b}^{2}) }{(a - b)(b - c)(a - c)} \\ \\ \sf = \frac{ {b}^{2} - {c}^{2} - {a}^{2} + {c}^{2} + {a}^{2} - {b}^{2} }{(a - b)(b - c)(a - c)} \\ \\ = \frac{0}{(a - b)(b - c)(a - c)} \\ \\\sf = > \: \: infinity \: \: \: \: \: \: ..answer \\ \\ \sf we \: know \: that \: = > \frac{0}{something} \: = \: infinity \: (always)\end{gathered}$

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