Question

If `a`, `b` and `c` are rational numbers and `a+b+c=0`, then prove that the roots of the equation `a(a^2-bc)x^2+b(b^2-ca)x+c(c^2-ab)=0` are also rational.?

Answer 1

If non-degenerate, then its roots are `x=c/a` and `x=1`

Explanation:

Given:

`a+b+c = 0`

We have:

`b = -a-c`

So:

`{(a(a^2-bc) = a(a^2-(-a-c)c) = a(a^2+ac+c^2)), (b(b^2-ca) = (-a-c)((-a-c)^2-ca) = (-a-c)(a^2+ac+c^2)), (c(c^2-ab) = c(c^2-a(-a-c)) = c(a^2+ac+c^2)) :}`

So either `a^2+ac+c^2 = 0`, in which case the given polynomial has all coefficients `0` and hence any value of `x` is a root, or we can divide all of the coefficients by `a^2+ac+c^2` to get:

`0 = ax^2-(a+c)x+c = (ax^2-ax)-(cx-c) = (ax-c)(x-1)`

If `a = 0`, then this is a linear equation with one root `x=1`.

If `a != 0` then this is a quadratic with roots `x=c/a` and `x=1`, both of which are rational.