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# If $a{x^2} + bx + c = 0$ and $b{x^2} + cx + a = 0$ have a common root and $abc \ne 0$, then $\dfrac{{{a^3} + {b^3} + {c^3}}}{{abc}} =$.A.3B.2C.1D.4

Last updated date: 13th Jun 2024
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Hint- As this question is of the concept of finding the roots, so it should be known that the sum of the roots is given by $- \dfrac{{{\text{coefficient}}\,{\text{of}}\,{\text{x}}}}{{{\text{coefficient}}\,{\text{of}}\,{{\text{x}}^2}}}$ and the product of roots is given by $- \dfrac{{{\text{constant}}}}{{{\text{coefficient}}\,{\text{of}}\,{{\text{x}}^2}}}$, and the in this case sum of the roots in both the equation will be equal to each other and also the product of both the roots will equal for both the equations.

The given quadratic equations are $a{x^2} + bx + c = 0$ and $b{x^2} + cx + a = 0$.
We need to find the value of $\dfrac{{{a^3} + {b^3} + {c^3}}}{{abc}}$.
Let,
$a{x^2} + bx + c = 0$ ………(i)
$b{x^2} + cx + a = 0$ …..….(ii)
Let $\alpha$ and $\beta$ be the roots common for the above equation.
The sum of the roots is given by $- \dfrac{{{\text{coefficient}}\,{\text{of}}\,{\text{x}}}}{{{\text{coefficient}}\,{\text{of}}\,{{\text{x}}^2}}}$ and the product of roots is given by $- \dfrac{{{\text{constant}}}}{{{\text{coefficient}}\,{\text{of}}\,{{\text{x}}^2}}}$.
Therefore,
For eq. (i), $\alpha + \beta = \dfrac{{ - b}}{a}$ and $\alpha \beta = \dfrac{c}{a}$.
For eq. (ii), $\alpha + \beta = \dfrac{{ - c}}{b}$ and $\alpha \beta = \dfrac{a}{b}$.
As both the roots are common, the sum and the product of the root will be equal for both the equations.
Therefore, it can be interpreted as follows:
$\dfrac{{ - b}}{a} = \dfrac{{ - c}}{b} \Rightarrow {b^2} = ac$ …….(iii)
$\Rightarrow c = \dfrac{{{b^2}}}{a}$ ……(iv)
$\Rightarrow a = \dfrac{{{b^2}}}{c}$ ……(v)
Again,
$\dfrac{c}{a} = \dfrac{a}{b} \Rightarrow {a^2} = bc$ ……(vi)
Now, use equation (iv) in equation (vi).
${a^2} = bc \\ {a^2} = b\left( {\dfrac{{{b^2}}}{a}} \right)\,\,\,\,\left[ {\because c = \dfrac{{{b^2}}}{a}} \right] \\ {a^2} = \dfrac{{{b^3}}}{a} \\ {a^2}\left( a \right) = {b^3} \\ {a^3} = {b^3} \\$
Now, use equation (v) in equation (vi).
${a^2} = bc \\ {\left( {\dfrac{{{b^2}}}{c}} \right)^2} = bc\,\,\,\left[ {\because a = \dfrac{{{b^2}}}{c}} \right] \\ \dfrac{{{b^4}}}{{{c^2}}} = bc \\ \dfrac{{{b^4}}}{b} = c\left( {{c^2}} \right) \\ {b^3} = {c^3} \\$
So, it can be observed that,
${a^3} = {b^3} = {c^3} \Rightarrow a = b = c$.
Now, use these values to find the value of $\dfrac{{{a^3} + {b^3} + {c^3}}}{{abc}}$.
$\dfrac{{{a^3} + {b^3} + {c^3}}}{{abc}} = \dfrac{{{a^3} + {a^3} + {a^3}}}{{a \cdot a \cdot a}} = \dfrac{{3{a^3}}}{{{a^3}}} = 3$
So,option A is the right answer

Note- The standard form of any quadratic equation is, $a{x^2} + bx + c = 0$ here x is the variable and a, b and c are the constants and provided $a \ne 0$.
Here is the case of two quadratic having same roots and to solve the equation let it be equal to some constant: $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}} = \lambda \left( {{\text{say}}} \right)$.
Here, ${a_1}$ and ${a_2}$ are the coefficient of ${x^2}$, ${b_1}$ and ${b_2}$ are the coefficient of $x$ and ${c_1}$ and ${c_2}$ are the constants of the equation.