Question

b+c/(a-b) (a-c) +c+a/(b-a) (b-c) +a+b/(c-a) (c-b) It solved this answer

Answer 1

☆Hey friend!!!!☆

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Here is your answer ☞
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$\frac{(b + c)}{(a - b)(a - c)} + \frac{(c + a)}{(b - a)(b - c)} + \frac{a + b}{(c - a)(c - b)} \\ \\ = \frac{(b + c)}{(a - b)(a - c)} - \frac{(a + c)}{(a - b)(b - c)} + \frac{(a + b)}{(a - c)(b - c)} \\ \\ = \frac{(b + c)(b - c) - (a + c)(a - c) + (a + b)(a - b)}{(a - b)(b - c)(a - c)} \\ \\ = \frac{( {b}^{2} - {c}^{2} ) - ( {a}^{2} - {c}^{2}) + ( {a}^{2} - {b}^{2} ) }{(a - b)(b - c)(a - c)} \\ \\ = \frac{ {b}^{2} - {c}^{2} - {a}^{2} + {c}^{2} + {a}^{2} - {b}^{2} }{(a - b)(b - c)(a - c)} \\ \\ = \frac{0}{(a - b)(b - c)(a - b)} \\ \\ \\ = Infinity \: \: \: \: .......answer \\ \\ we \: know \: that \: = > \frac{0}{something} \: is \: always \: eqaule \: to \: "Infinity "$

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Devil_king ▄︻̷̿┻̿═━一

Answer 2

Explanation:

ʏᴏᴜ ᴄᴀɴ sᴇᴇ ᴛʜᴇʀᴇ ɪᴛ's ᴀᴠᴀɪʟᴀʙʟᴇ.........