Question

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

Answer 1

Hint:
It is given in the question that if $a{x^2} + bx + c = o$ and $b{x^2} + cx + a = o$ have a common root and $abc \ne 0$ .
Then what is the value of $\dfrac{{{a^3} + {b^3} + {c^3}}}{{abc}}$ .
Firstly, we will find the roots of the given equations. Then after, by using $\dfrac{{{x^2}}}{{{r_2}{q_1} - {r_1}{q_2}}} = \dfrac{{ - x}}{{{p_1}{r_2} - {r_1}{p_2}}} = \dfrac{1}{{{p_1}{q_2} - {q_1}{p_2}}}$ we will find the values of x.
Finally, we will compare the values of x and we will get the answer.

Complete step by step solution:
It is given in the question that if $a{x^2} + bx + c = o$ and $b{x^2} + cx + a = o$ have a common root and $abc \ne 0$.
Then what is the value of $\dfrac{{{a^3} + {b^3} + {c^3}}}{{abc}}$ .
Since, the common roots of $a{x^2} + bx + c = o$ and $b{x^2} + cx + a = o$ are
 ${p_1}{x^2} + {q_1}x + {r_1} = 0$ and ${p_2}{x^2} + {q_2}x + {r_2} = 0$
 $\therefore \dfrac{{{x^2}}}{{{r_2}{q_1} - {r_1}{q_2}}} = \dfrac{{ - x}}{{{p_1}{r_2} - {r_1}{p_2}}} = \dfrac{1}{{{p_1}{q_2} - {q_1}{p_2}}}$
Since, $a{x^2} + bx + c = o$ and $b{x^2} + cx + a = o$
Therefore, the values of ${p_1},{p_2},{q_1},{q_2},{r_1},{r_2}$ are as follows:
  $
\Rightarrow {p_1} = a,{p_2} = b \\
\Rightarrow {q_1} = b,{q_2} = c \\
\Rightarrow {r_1} = c,{r_2} = a \\
 $
Now, put the values of ${p_1},{p_2},{q_1},{q_2},{r_1},{r_2}$ in $\dfrac{{{x^2}}}{{{r_2}{q_1} - {r_1}{q_2}}} = \dfrac{{ - x}}{{{p_1}{r_2} - {r_1}{p_2}}} = \dfrac{1}{{{p_1}{q_2} - {q_1}{p_2}}}$ .
 $\therefore \dfrac{{{x^2}}}{{ab - {c^2}}} = \dfrac{{ - x}}{{{a^2} - bc}} = \dfrac{1}{{ac - {b^2}}}$
Now, by comparing $\dfrac{{ - x}}{{{a^2} - bc}} = \dfrac{1}{{ac - {b^2}}}$ , we get,
 $\Rightarrow x = \dfrac{{bc - {a^2}}}{{ac - {b^2}}}$ (I)
Similarly, by comparing $\dfrac{{{x^2}}}{{ab - {c^2}}} = \dfrac{{ - x}}{{{a^2} - bc}}$ , we get,
 $\Rightarrow x = \dfrac{{{c^2} - ab}}{{{a^2} - bc}}$ (II)
Now, compare equation (I) and (II)
 $\Rightarrow \dfrac{{bc - {a^2}}}{{ac - {b^2}}} = \dfrac{{{c^2} - ab}}{{{a^2} - bc}}$
 \[\Rightarrow \left( {bc - {a^2}} \right)\left( {{a^2} - bc} \right) = \left( {{c^2} - ab} \right)\left( {ac - {b^2}} \right)\]
 \[\Rightarrow {a^2}bc - {a^4} - {b^2}{c^2} + {a^2}bc = a{c^3} - {b^2}{c^2} - {a^2}bc + a{b^3}\]
 \[{a^4} + a{b^3} + a{c^3} = 3{a^2}bc\]
Now, take out ‘a’ common from the L.H.S
 \[\Rightarrow a\left( {{a^3} + {b^3} + {c^3}} \right) = 3{a^2}bc\]
 \[\Rightarrow \dfrac{{a\left( {{a^3} + {b^3} + {c^3}} \right)}}{{3{a^2}bc}} = 1\]
 \[\Rightarrow \therefore \dfrac{{\left( {{a^3} + {b^3} + {c^3}} \right)}}{{abc}} = 3\]

Therefore, the value of \[\dfrac{{\left( {{a^3} + {b^3} + {c^3}} \right)}}{{abc}} = 3\]

Note:
The formula of ${\left( {a + b + c} \right)^3}$ can be proven as the following method
Prove ${\left( {a + b + c} \right)^3}$
 ${\left( {a + b + c} \right)^3} = {\left[ {\left( {a + b} \right) + c} \right]^3}$
 ${\left( {a + b + c} \right)^3} = {\left( {a + b} \right)^3} + 3{\left( {a + b} \right)^2}c + 3\left( {a + b} \right){c^2} + {c^3}$
Now, using the formula of ${\left( {a + b} \right)^2},$ we get,

 $\Rightarrow {\left( {a + b + c} \right)^3} = \left( {{a^3} + 3{a^2}b + 3a{b^2} + {b^3}} \right) + 3\left( {{a^2} + 2ab + {b^2}} \right)c + 3\left( {a + b} \right){c^2} + {c^3}$
 $\Rightarrow {\left( {a + b + c} \right)^3} = {a^3} + {b^3} + {c^3} + 3{a^2}b + 3{a^2}c + 3a{b^2} + 3{b^2}c + 3a{c^2} + 3b{c^2} + 6abc$
 $\Rightarrow {\left( {a + b + c} \right)^3} = {a^3} + {b^3} + {c^3} + 3{a^2}b + 3{a^2}c + 3a{b^2} + 3{b^2}c + 3a{c^2} + 3b{c^2} + 3abc + 3abc$
 $\Rightarrow {\left( {a + b + c} \right)^3} = {a^3} + {b^3} + {c^3} + 3a\left( {ab + ac + {b^2} + bc} \right) + 3c\left( {ab + ac + {b^2} + bc} \right)$
 $\Rightarrow {\left( {a + b + c} \right)^3} = {a^3} + {b^3} + {c^3} + 3\left( {a + c} \right)\left( {ab + ac + {b^2} + bc} \right)$
 $\Rightarrow {\left( {a + b + c} \right)^3} = {a^3} + {b^3} + {c^3} + 3\left( {a + c} \right)\left[ {a\left( {b + c} \right) + b\left( {b + c} \right)} \right]$
 $\Rightarrow {\left( {a + b + c} \right)^3} = {a^3} + {b^3} + {c^3} + 3\left( {a + c} \right)\left( {a + b} \right)\left( {b + c} \right)$
Hence, ${\left( {a + b + c} \right)^3} = {a^3} + {b^3} + {c^3} + 3\left( {a + c} \right)\left( {a + b} \right)\left( {b + c} \right)$ .