Let `a, b, c, d` be real numbers in `G.P.` If `u, v, w` satisfy the system of equations `u + 2y +3w = 6,4u + 5v + 6w =12 and 6u + 9v = 4` then show that the roots of the equation `(1/u+1/v+/w)x^2+[(b-c)^2+(c-a)^2+(d-b)^2]x+u+v+w=0` and 20x^2+10(a-d)^2 x-9=0` are reciprocals of each other.

Options

A. `a-d`
B. `(a-d)^(2)`
C. `a^(2)-d^(2)`
D. `(a+d)^(2)`

Explanation

Let `b=ar,c=ar^(2)` and `d=ar^(3)`
Now `(b-c)^(2)+(c-a)^(2)+(d-b)^(2)`
`=(ar-ar^(2))^(2)+(ar^(2)-a)^(2)+(ar^(3)-ar)^(2)`
`=a^(2)r^(2)(1-r)^(2)+a^(2)(r^(2)-1)^(2)+a^(2)r^(2)(r^(2)-1)^(2)`
`=a^(2)(1-r)^(2){r^(2)+(r+1)^(2)+r^(2)(r+1)^(2)}`
`=a^(2)(1-r)^(2)(r^(4)+2r^(3)+3r^(2)+2r+1)`
`=a^(2)(1-r)^(2)(1+r+r^(2))^(2)=a^(2)(1-r^(3))^(2)`
`=(a-ar^(3))^(2)=(a-d)^(2)`