Question

Let \[\alpha ,\beta ,\gamma \] be the roots of the cubic equation \[{a_0}{x^3} + 3{a_1}{x^2} + 3{a_2}x + {a_3} = 0\left( {{a_0} \ne 0} \right)\]. Then the value of \[{\left( {\alpha - \beta } \right)^2} + {\left( {\beta - \gamma } \right)^2} + {\left( {\gamma - \alpha } \right)^2}\] equals
A) \[\dfrac{{18\left( {{a_2}^2 - {a_0}{a_1}} \right)}}{{{a_0}^2}}\]
B) \[\dfrac{{18\left( {{a_2}^2 + {a_0}{a_1}} \right)}}{{{a_0}^2}}\]
C) \[\dfrac{{18\left( {{a_0}^2 - {a_1}{a_2}} \right)}}{{{a_0}^2}}\]
D) \[\dfrac{{18\left( {{a_1}^2 - {a_0}{a_2}} \right)}}{{{a_0}^2}}\]

Answer 1

Hint:
Here we will simply write the basic conditions of the roots of the cubic equation. Then we will expand the given equation and solve it. Then we will put the value of the roots conditions in the equation to get the value of the given equation.

Complete step by step solution:
Given cubic equation is \[{a_0}{x^3} + 3{a_1}{x^2} + 3{a_2}x + {a_3} = 0\left( {{a_0} \ne 0} \right)\].
It is given that \[\alpha ,\beta ,\gamma \] are the roots of the given cubic equation.
We know the three basic conditions of the roots of the cubic equation. Therefore, we get
\[\begin{array}{l}
\alpha + \beta + \gamma = \dfrac{{ - 3{a_1}}}{{{a_0}}}\\
\alpha \beta + \beta \gamma + \alpha \gamma = \dfrac{{3{a_2}}}{{{a_0}}}\\
\alpha \beta \gamma = \dfrac{{ - {a_3}}}{{{a_0}}}
\end{array}\]
Now we have to find the value of \[{\left( {\alpha - \beta } \right)^2} + {\left( {\beta - \gamma } \right)^2} + {\left( {\gamma - \alpha } \right)^2}\]. So we will simply expand this equation by opening the square of the terms. Therefore, we get
\[ \Rightarrow {\left( {\alpha - \beta } \right)^2} + {\left( {\beta - \gamma } \right)^2} + {\left( {\gamma - \alpha } \right)^2} = \left( {{\alpha ^2} + {\beta ^2} - 2\alpha \beta } \right) + \left( {{\beta ^2} + {\gamma ^2} - 2\beta \gamma } \right) + \left( {{\alpha ^2} + {\gamma ^2} - 2\alpha \gamma } \right)\]
\[ \Rightarrow {\left( {\alpha - \beta } \right)^2} + {\left( {\beta - \gamma } \right)^2} + {\left( {\gamma - \alpha } \right)^2} = 2\left( {{\alpha ^2} + {\beta ^2} + {\gamma ^2}} \right) - 2\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right)\]
We know that \[{\left( {\alpha + \beta + \gamma } \right)^2} = {\alpha ^2} + {\beta ^2} + {\gamma ^2} + 2\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right)\]. By this we will get the value of \[{\alpha ^2} + {\beta ^2} + {\gamma ^2} = {\left( {\alpha + \beta + \gamma } \right)^2} - 2\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right)\]. Therefore, y putting this value in the above equation we get
\[ \Rightarrow {\left( {\alpha - \beta } \right)^2} + {\left( {\beta - \gamma } \right)^2} + {\left( {\gamma - \alpha } \right)^2} = 2\left( {{{\left( {\alpha + \beta + \gamma } \right)}^2} - 2\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right)} \right) - 2\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right)\]
\[ \Rightarrow {\left( {\alpha - \beta } \right)^2} + {\left( {\beta - \gamma } \right)^2} + {\left( {\gamma - \alpha } \right)^2} = 2{\left( {\alpha + \beta + \gamma } \right)^2} - 4\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right) - 2\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right)\]
\[ \Rightarrow {\left( {\alpha - \beta } \right)^2} + {\left( {\beta - \gamma } \right)^2} + {\left( {\gamma - \alpha } \right)^2} = 2{\left( {\alpha + \beta + \gamma } \right)^2} - 6\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right)\]
Now we will put the value of the \[\left( {\alpha + \beta + \gamma } \right)\] and \[\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right)\] in the above equation, we get
\[ \Rightarrow {\left( {\alpha - \beta } \right)^2} + {\left( {\beta - \gamma } \right)^2} + {\left( {\gamma - \alpha } \right)^2} = 2{\left( {\dfrac{{ - 3{a_1}}}{{{a_0}}}} \right)^2} - 6\left( {\dfrac{{3{a_2}}}{{{a_0}}}} \right)\]
Now we will solve this equation, we get
\[ \Rightarrow {\left( {\alpha - \beta } \right)^2} + {\left( {\beta - \gamma } \right)^2} + {\left( {\gamma - \alpha } \right)^2} = 2\left( {\dfrac{{9{a_1}^2}}{{{a_0}^2}}} \right) - \left( {\dfrac{{18{a_2}}}{{{a_0}}}} \right)\]
\[ \Rightarrow {\left( {\alpha - \beta } \right)^2} + {\left( {\beta - \gamma } \right)^2} + {\left( {\gamma - \alpha } \right)^2} = \left( {\dfrac{{18{a_1}^2}}{{{a_0}^2}}} \right) - \left( {\dfrac{{18{a_2}}}{{{a_0}}}} \right)\]
Now we will take \[\dfrac{{18}}{{{a_0}^2}}\] common from both the terms. Therefore, we get
\[ \Rightarrow {\left( {\alpha - \beta } \right)^2} + {\left( {\beta - \gamma } \right)^2} + {\left( {\gamma - \alpha } \right)^2} = \dfrac{{18}}{{{a_0}^2}}\left( {{a_1}^2 - {a_0}{a_2}} \right) = \dfrac{{18\left( {{a_1}^2 - {a_0}{a_2}} \right)}}{{{a_0}^2}}\]
Hence, the value of \[{\left( {\alpha - \beta } \right)^2} + {\left( {\beta - \gamma } \right)^2} + {\left( {\gamma - \alpha } \right)^2}\] is equals to \[\dfrac{{18\left( {{a_1}^2 - {a_0}{a_2}} \right)}}{{{a_0}^2}}\].

So, option D is the correct option.

Note:
Here we should note that the cubic equations are the equation in which the highest exponent of the variable is 3. In an equation the number of its roots is equal to the value of the highest exponent of the variable of that equation. That means for a cubic equation there are three roots of the equation. Apart from the cubic equations, equations are also categorized as quadratic equations, linear equations and so on. A quadratic equation is an equation in which the highest exponent of the variable is 2 whereas a linear equation has the highest degree of 1. So there are only two roots of the quadratic equation and one in the linear equation.
Representation of equations is done in the following manner:
Linear equation: \[ax+b=0\]
Quadratic equation: \[a{x^2} + bx + c = 0\]
Cubic equation: \[a{x^3} + b{x^2} + cx + d = 0\]