Manipulating the Roots of a Cubic

Let x =(1+a)/(1-a) .. By componendo and dividendo we get a=(x-1)/(x+1) Since 'a' is root of given equation so substituting the value of 'a' in given equation that equation transforms to an equation whose roots are (1+a)/(1-a) , (1+b)/(1-b) , (1+c)/(1-c) Now find sum of roots ofobtaining equation using formula -(coeff of x^2)/(coeff of x^3) We get required result

Let's work on the fractions $(\frac{1+a}{1-a}+1)+(\frac{1+b}{1-b}+1)+(\frac{1+c}{1-c}+1)-3$ $=(\frac{1+a}{1-a}+\frac{1-a}{1-a})+(\frac{1+b}{1-b}+\frac{1-b}{1-b})+(\frac{1+c}{1-c}+\frac{1-c}{1-c})-3$ $2(\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c})-3,$ so $2(\frac{2-2(a+b+c)+(ab+ac+bc)+1}{(1-a)(1-b)(1-c)})-3$ $2(\frac{2-2(a+b+c)+(ab+ac+bc)+1}{1-(a+b+c)+ab+ac+bc-abc})-3,$ ok now by Vieta's formulas we have $a+b+c=0$ $ab+ac+bc=1$ $abc=1,$ hence $2\frac{2+0+1+1}{1-0+1-1}-3=8-3=5,$ so $\boxed{\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}=5}$

P is the correct answer. We have: P=(3-a-b-c-ab-bc-ac+3abc) / (1-a-b-c+ab+ac+bc-abc); Apply Viet theorem, we have: a+b+c=0; ab+ac+bc=1; abc=1; Hence P=5/1=5. 5 is the correct answer.