For the following two (02) items: Consider the equation abx^2 + bcx...

The given equation is $abx^2 + bcx + ca = cax^2 + abx + bc$.

Rearranging the terms, we get:

$(ab - ca)x^2 + (bc - ab)x + (ca - bc) = 0$

$a(b - c)x^2 + b(c - a)x + c(a - b) = 0$

The sum of the coefficients is:

$a(b - c) + b(c - a) + c(a - b) = ab - ac + bc - ab + ac - bc = 0$

Since the sum of the coefficients is zero, one root of the equation is $x = 1$.

It is given that the roots of the equation are equal, so the other root must also be $x = 1$.

The product of the roots is given by $\dfrac{c(a - b)}{a(b - c)}$.

Equating the product of the roots to $1 \times 1 = 1$:

$\dfrac{c(a - b)}{a(b - c)} = 1$

$ca - cb = ab - ac$

$2ca = ab + bc$

Dividing both sides by $abc$, we get:

$\dfrac{2}{b} = \dfrac{1}{c} + \dfrac{1}{a}$

Answer: $\dfrac{1}{a} + \dfrac{1}{c} = \dfrac{2}{b}$