Let a,b,c be different nonzero real numbers and x,y,z be three numbers satisfying the system of equations
`(x)/(a)+(y)/(a-1)+(z)/(a+1)=1`
`(x)/(b)+(y)/(b-1)+(z)/(b+1)=1`
`(x)/(c)+(y)/(c-1)+(z)/(c+1)=1`
Then which of the following is correct?

Options

A. `x+y+z=a+b+c`
B. `x=-abc`
C. `z=((1+a)(a+b)(a+c))/(2)`
D. `y=((1-a)(a-b)(a-c))/(2)`

Answer

A, B, D

Explanation

Equation
`(x)/(t)+(y)/(t-1)+(z)/(t+1)=1` has three
roots a,b,c
by solving the equation we get
`(x(t^(2)-1)+y(t^(2)+t)+z(t^(2)-t))/(t(^(2)-1))=1`
`t^(3)-t=(x+y+z)t^(2)+(y-z)t-x`
`t^(3)(x+y+z)t^(2)+(-y+z-1)t+x=`
`abc=-x` .(1)
`a+b+c=x+y+z` ...(2)
`ab+bc+ca=-y+z-1` ...(3)
`x=-abc`
`y+z=a+b+c+abc`
`y-z=-1-ab-bc-ca`
`2y=a+b+c+abc-1-ab-bc-ca`
`2y=(a-1)(b-1)(c-1)`
`y=((a-1)(b-1)(c-1))/(2)`
`2z=1+a+b+c+ab+bc+ca+abc`
`2z(1+a)(1+b)(1+c)`
`z=((1+a)(1+b)(1+c))/(2)`
so (C) is incorrect option