If the points `((a^3)/((a-1))),(((a^2-3))/((a-1))),((b^3)/(b-1)),(((b^2-3)/((b-1))),` and `(((c^2-3))/((c-1))),` where `a , b , c` are different from 1, lie on the `l x+m y+n=0` , then

Options

A. (1) `a+b+c = -(m)/(l)`
B. (2) `ab+bc+ca=(n)/(l)`
C. (3) `abc = ((m+n))/(l)`
D. (4) abc-(bc+ca+ab) +3(a+b+c)=0

Answer

A, B, D

Explanation

Since the given points lie on the line lx+my+n=0, a,b,c are the roots of the equation
`l((t^(3))/(t-1)) + m((t^(3)-3)/(t-1))+n=0`
`"or " l t^(3) +mt^(2) + nt-(3m+n) = 0 " " (1)`
`"Hence, "a+b+c = -(m)/(l)`
`ab+bc+ca = (n)/(l)" " (2)`
`abc = (3m+n)/(l) " " (3)`
So, from (1), (2), and (3), we get
abc-(bc+ca+ab)+3(a+b+c)=0`