Question

The common roots of the equation ${x^3} + 2{x^2} + 2x + 1 = 0$ and ${x^{1988}} + {x^{130}} + 1 = 0$ are (where $\omega $ is a non real cube of unity).
1) $\omega $
2) ${\omega ^2}$
3) $ - 1$
4) $\omega - {\omega ^2}$

Answer 1

Hint: Factorise the equation ${x^3} + 2{x^2} + 2x + 1 = 0$ and equate the factors to 0 to find the roots of ${x^3} + 2{x^2} + 2x + 1 = 0$. Substitute the roots in the second equation to check whether it is the common root of both the equations or not.

Complete step-by-step answer:
Find the roots of ${x^3} + 2{x^2} + 2x + 1 = 0$
Factorise the given polynomial, ${x^3} + 2{x^2} + 2x + 1 = 0$
$
  {x^3} + 2{x^2} + 2x + 1 = 0 \\
  \left( {{x^3} + 1} \right) + 2x\left( {x + 1} \right) = 0 \\
  \left( {x + 1} \right)\left( {{x^2} - x + 1} \right) + 2x\left( {x + 1} \right) = 0 \\
$
Take $\left( {x + 1} \right)$ common.
$
  \left( {x + 1} \right)\left( {\left( {{x^2} - x + 1} \right) + 2x} \right) = 0 \\
  \left( {x + 1} \right)\left( {{x^2} + x + 1} \right) = 0 \\
 $
Equate each factor to 0.
$
  \left( {x + 1} \right) = 0 \Rightarrow x = - 1 \\
  \left( {{x^2} + x + 1} \right) = 0 \Rightarrow x = \dfrac{{ - 1 \pm \sqrt { - 3} }}{2} \\
 $
The value $\dfrac{{ - 1 + \sqrt { - 3} }}{2}$ is known to be equal to the $\omega $, the non real cube root of the unity. $\dfrac{{ - 1 - \sqrt { - 3} }}{2}$ is known to be ${\omega ^2}$.
Here, $\omega $ and ${\omega ^2}$ are the roots of unity.
But, $x = - 1$ does not satisfy the equation, ${x^{1988}} + {x^{130}} + 1 = 0$as
${\left( { - 1} \right)^{1988}} + {\left( { - 1} \right)^{130}} + 1 = 1 - 1 + 1 = 1 \ne 0$
Substitute both the roots in the equation ${x^{1988}} + {x^{130}} + 1 = 0$
$
  {\left( \omega \right)^{1988}} + {\left( \omega \right)^{130}} + 1 = 0 \\
  1.{\omega ^2} + 1.\omega + 1 = 0 \\
$
As, $\omega $ and ${\omega ^2}$ are the roots of unity, therefore, ${\omega ^2} + \omega + 1 = 0$.
Similarly,
$
  {\left( {{\omega ^2}} \right)^{1988}} + {\left( {{\omega ^2}} \right)^{130}} + 1 = 0 \\
  1.\omega + 1.{\omega ^2} + 1 = 0 \\
$
Hence, common roots of equation ${x^3} + 2{x^2} + 2x + 1 = 0$ and ${x^{1988}} + {x^{130}} + 1 = 0$ are $\omega $ and ${\omega ^2}$.
Therefore, option A and B are correct.

Note: The root of an equation is the value of $x$ for which the equation is equal to 0. As, $\omega $ and ${\omega ^2}$ are the roots of unity, therefore, ${\omega ^2} + \omega + 1 = 0$. The value of $\omega $ is equal to $\dfrac{{ - 1 + \sqrt { - 3} }}{2}$.