please answer this
no spam or I will report your account
Answers
Given relation,
Multiplying both sides by (a + b + c) , [(a + b + c) ≠ 0]
Hence,
Hence, Proved ✔️
I hope it helps you ❤️✔️
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
______________________________________