If `alpha, beta, gamma` are the roots of the cubic equation `x^(3)+qx+r=0` then the find equation whose roots are `(alpha-beta)^(2),(beta-gamma)^(2),(gamma-alpha)^(2)`.

Explanation

`:' alpha, beta, gamma` are the roots of the cubic equation
`x^(3)+qx+r=0`………i
Then `sum alpha=0, sum alpha beta=q, alpha beta gamma =-r`……..ii
If y is a root of the required equation, then
`y=(alpha-beta)^(2)=(alpha+beta)^(2)-4 alpha beta`
`=(alpha+beta+gamma-gamma)^(2)-(4 alpha beta gamma)/(gamma)`

`=(0-gamma)^(2)+(4r)/(gamma)` [from Eq. (ii) ]`
`impliesy=gamma^(2)+(4r)/(gamma)`
[replacing `gamma` by `x` which is a root of Eq. (i) ]
`:.y=x^(2)+(4r)/x`
Or `x^(3)-yx+4r=0`..........iii
The required equation is obatained by eliminating `x` between Eqs (i) and (iii)
Now, subtracting Eq. (iii) from Eq. (i) we get
`(q+y)x-3r=0`
or `x=(3r)/(q+y)`
On substituting the value of `x` in Eq. (i) we get
`((3r)/(q+y))^(3)+q((3r)/(q+y))+r=0`
Thus, `y^(3)+6qy^(2)+9q^(2)y+(4q^(3)+27r^(2))=0`
which is the required equation.