If one root of a(x^2)+bx+c=0 is the square of the other,then which of the following is true-options below?

Let the roots be p and p^2.

p + p^2 = -b/a

(p) (p^2) = c/a, or, p^3 = c/a

Now, 1 - p^3 = (1-p) (1+p+p^2)

Or, 1 - (c/a) = (1-p) (1 - (b/a)).

Or, (a-c)/a = (1-p) (a-b)/a

Or, (a-c) = (1-p) (a-b)

Or, (a-c) / (a-b) = (1-p)

p = 1 - (1-p) = 1 - [(a-c) / (a-b)]

= ((a-b) - (a-c)) / (a-b)

= (c-b) / (a-b).... equation (1)

Since p is a root of the given equation, it will satisfy it. So,

ap^2 + bp + c = 0

Or, a (a-b)^2 p^2 + b (a-b)^2 p + c (a-b)^2 = 0

Or, a (c-b)^2 + b (c-b) (a-b) + c (a-b)^2 = 0 [by substituting the value of p using equation (1)]

Or, a (c^2 - 2bc + b^2) + b (ca - ab - bc + b^2) + c (a^2 - 2ab + b^2) = 0

Or, ac^2 - 2abc + ab^2 + abc - ab^2 - cb^2 + b^3 + ca^2 - 2abc + cb^2 = 0

Or, ac^2 + b^3 + ca^2 = 3abc

Hence, option C is correct.

Let "λ" be a root of the quadratic.

The quadratic we are looking for is:

(x - λ)(x - λ²) = 0

x² - λ²x - λx + λ³ = 0

x² + (-λ² - λ)x + λ³ = 0

a = 1

b = -λ² - λ

c = λ³

3abc = 3(1)(-λ² - λ)(λ³)

3abc = -3λ⁵ - 3λ⁴

That means the left side has to be of a power 5. The only answer that does this is D).

EDIT: Nope, I was wrong. It is in fact option C).