Question

Please i cannot slove it. can u​

Answer 1

Explanation:

`a^(-1)=(1)/(a) \ ; \ b^(-1)=(1)/(b) \ ; \ c^(-1)=(1)/(c)`

`(1)/(1+a+b^(-1))+(1)/(1+b+c^(-1))+(1)/(1+c+a^(-1))=(1)/(1+a+ (1)/(b))+(1)/(1+b+(1)/(c))+(1)/(1+c+(1)/(a))`

`= (1)/((1*b)/(1*b)+(a*b)/(1*b)+(1)/(b))+(1)/((1*c)/(1*c)+(b*c)/(1*c)+(1)/(c))+(1)/((1*a)/(1*a)+(c*a)/(1*a)+(1)/(a))\\\\\\=(1)/((b+ab+1)/(b))+(1)/((c+bc+1)/(c))+(1)/((a+ac+1)/(a))\\\\\\=(b)/(b+ab+1)+(c)/(c+bc+1)+(a)/(a+ac+1)`

abc = 1

c = 1/ab

`=(b)/(b+ab+1)+((1)/(ab))/((1)/(ab)+b*(1)/(ab)+1)+(a)/(a+a*(1)/(ab)+1)\\\\\\`

`= (b)/(b+ab+1)+((1)/(ab))/((1+b+ab)/(ab))+(a)/((ab+1+b)/(b))\\\\\\=(b)/(b+ab+1)+((1)/(ab)*ab)/(1+b+ab)+(ab)/(ab+1+b)`

`=(b)/(b+ab+1)+(1)/(1+b+ab)+(ab)/(1+b+ab)\\\\\\=(b+1+ab)/(b+1+ab)\\\\\\=1`