Question

If one root of $a{x^2} + bx + c = 0$ be square of the other, then the value of ${b^3} + a{c^2} + {a^2}c$ is:
A. $3abc$
B. $ - 3abc$
C. $0$
D. None of these

Answer 1

Hint: One or more algebraic terms that each consist of a constant multiplied by one or more variables raised to a positive integral power is called polynomials in mathematics.
We need to use the relationship between the sum and product of the roots with the coefficients and the constant term to solve the given question.

Formula used:
The roots of a quadratic equation $a{x^2} + bx + c$ are $\alpha ,\beta $:
The sum of the roots is$(\alpha + \beta ) = \dfrac{{ - b}}{a}$,
The products of the roots is $\alpha \beta = \dfrac{c}{a}$.
And, the algebraic identity which is used in the question to expand the equation:
${(x + y)^3} = {x^3} + {y^3} + 3xy(x + y)$

Complete step by step Solution:
In the question, the equation is given by:
$a{x^2} + bx + c = 0$
Let $\alpha $and ${\alpha ^2}$be the roots of the given equation,
As we know that the sum of the roots of the equation is:$(\alpha + \beta ) = \dfrac{{ - b}}{a}$, then we have:
$\alpha + {\alpha ^2} = \dfrac{{ - b}}{a}\,\,\,\,....(i)$
Similarly, the product of the roots of the equation is:$\alpha \beta = \dfrac{c}{a}$, we have:
$\alpha \cdot {\alpha ^2} = \dfrac{c}{a}\,\,\,\,....(ii)$
Now, consider the equation $(i)$, cubing both sides of the equation, then:
${\left( {\alpha + {\alpha ^2}} \right)^3} = {\left( {\dfrac{{ - b}}{a}} \right)^3}$
Using the algebraic identity of cube ${(x + y)^3} = {x^3} + {y^3} + 3xy(x + y)$to expand the above equation, so:
${\alpha ^3} + {\alpha ^6} + 3\alpha \cdot {\alpha ^2}(\alpha + {\alpha ^2}) = {\left( {\dfrac{{ - b}}{a}} \right)^3} \\$
$\Rightarrow {\alpha ^3} + {\alpha ^6} + 3{\alpha ^3}(\alpha + {\alpha ^2}) = {\left( {\dfrac{{ - b}}{a}} \right)^3} \\$
Consider the equation $(ii)$, if we expand the equation then we have:
${\alpha ^3} = \dfrac{c}{a}$
Substitute the value of ${\alpha ^3}$and $\alpha + {\alpha ^2}$ in the obtained equation:
$\left( {\dfrac{c}{a}} \right) + {\left( {\dfrac{c}{a}} \right)^2} + 3\left( {\dfrac{c}{a}} \right)\left( {\dfrac{{ - b}}{a}} \right) = {\left( {\dfrac{{ - b}}{a}} \right)^3}$
Then, multiply the equation by ${a^3}$, we obtain:
$\left( {\dfrac{c}{a}} \right){\left( a \right)^3} + {\left( {\dfrac{c}{a}} \right)^2}{\left( a \right)^3} + 3\left( {\dfrac{c}{a}} \right)\left( {\dfrac{{ - b}}{a}} \right){\left( a \right)^3} = {\left( {\dfrac{{ - b}}{a}} \right)^3}{\left( a \right)^3} \\$
$\Rightarrow {a^2}c + a{c^2} - 3abc = {b^3} \\$
$\Rightarrow {b^3} + {a^2}c + a{c^2} = 3abc \\$

Therefore, the correct option is (A).

Note: It should be noted that different approaches can be used to solve this issue. In which the initial algebraic formula ${(x + y)^3} = {x^3} + {y^3} + 3xy(x + y)$ can be used directly, substituting the values. Utilize the remaining information after that, then simplify the equation to obtain the precise answer. But there will be some challenging calculations involved.