Math, asked by mamonimandalma84, 1 month ago

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Answered by kinzal
10

Given relation,

 \sf \longrightarrow \frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} = 1 \\

Multiplying both sides by (a + b + c) , [(a + b + c) ≠ 0]

Hence,

 \sf \longrightarrow \sf \frac{a}{b + c} (a + b + c) + \frac{b}{c + a} (a + b + c) + \frac{c}{a + b} (a + b + c) = (a + b + c) \\ \\

 \sf \longrightarrow  \bigg( \frac{a^{2}}{b + c} + \frac{a( b+ c)}{b +c} \bigg) + \bigg( \frac{b^{2}}{c + a} + \frac{b( c + a)}{c + a} \bigg) + \bigg( \frac{c^{2}}{a + b} + \frac{c ( a + b)}{a + b} \bigg) = (a + b + c) \\ \\

 \sf \longrightarrow  \bigg( \frac{a^{2}}{b + c} + \frac{a\cancel{( b+ c)}}{\cancel{b +c}} \bigg) + \bigg( \frac{b^{2}}{c + a} + \frac{b\cancel{( c + a)}}{\cancel{c + a}} \bigg) + \bigg( \frac{c^{2}}{a + b} + \frac{c \cancel{( a + b)}}{\cancel{a + b}} \bigg) = (a + b + c) \\ \\

 \sf \longrightarrow  \bigg( \frac{a^{2}}{b + c} +  a \bigg) + \bigg( \frac{b^{2}}{c + a} + b \bigg) + \bigg( \frac{c^{2}}{a + b} + c  \bigg) = (a + b + c) \\ \\

 \sf \longrightarrow \bigg( \frac{a^{2}}{b + c} \bigg) + \bigg( \frac{b^{2}}{c + a} \bigg) + \bigg( \frac{c^{2}}{a + b}  \bigg)  + (a + b + c)= (a + b + c)  \\ \\

 \sf \longrightarrow  \frac{ {a}^{2} }{b + c}  +  \frac{ {b}^{2} }{c + a}  +  \frac{ {c}^{2} }{a + b}  = (a +  b + c) - (a + b + c) \\ \\

 \sf \longrightarrow  \frac{ {a}^{2} }{b + c}  +  \frac{ {b}^{2} }{c + a}  +  \frac{ {c}^{2} }{a + b}  = 0 \\ \\

Hence, Proved ✔️

I hope it helps you ❤️✔️


Saby123: Nice
Answered by Saby123
13

Solution :

a/b+c + b/c+a + c/a+b = 1

>> [ a(a+b)(c+a) + b(b+c)(a+b) + c(b+c)(c+a) ]/(a+b)(b+c)(c+a) = 1

>> [ (a²+ab)(c+a) + (b²+bc)(a+b) + (c²+bc)(c+a) ]/(a+b)(b+c)(c+a) = 1

>> [ a²c + a³ + abc + a²b + ab² + b³ + abc + b²c + c³ +ac² + bc² + abc ]/(a+b)(b+c)(c+a) = 1

>> [ a³ + b³ + c³ + 3abc + ab(a+b) + bc(b+c) + ca(c+a) ]/(a+b)(b+c)(c+a) = 1

Transpose 1 to LHS

(a+b)(b+c)(a+c) = ab(a+b) + bc(b+c) + ca(c+a)

>> [a³+b³+c³+3abc]/(a+b)(b+c)(c+a) = 0

So , a³+b³+c³ + 3abc = 0 [ or a³+b³+c³ = -3abc ]

To prove :

a²/(b+c) + b²/(c+a) + c²/(a+b) = 0

LHS -

>> a²/(b+c) + b²/(c+a) + c²/(a+b)

>> [a²(c+a)(a+b) + b²(b+c)(a+b) + c²(b+c)(c+a) ]/(a+b)(b+c)(c+a)

>> [ (a+b+c)(a³+b³+c³+3abc) ]/(a+b)(b+c)(c+a)

>> 0

Hence Proved

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