Question

The common roots of the equation $x^3+2x^2+2x+1=0$ and $1+x^{130}+x^{1988}=0$ is/are
Note: $\omega$ is a non-real cube root of unity.

• A. $\omega$
• B. ${\omega }^2$
• C. $-1$
• D. $\omega -{\omega }^2$

Answer 1

Answer: (A), (B)

The correct options are
A $\omega$
B ${\omega }^2$

Let $f\left(x\right)=x^3+2x^2+2x+1$
As $f(-1)={(-1)}^3+2{(-1)}^2+2(-1)+1=0$
Then $x=-1$ is one root
By factorizing it we get,
$f\left(x\right)=(x+1)(x^2+x+1)=0$
And roots of $x^2+x+1$ are $w,w^2$
Now substituting $x=w$ in
$1+x^{130}+x^{1988}=1+w^{130}+w^{1988}=1+w^{129+1}+w^{1986+2}=1+w+w^2=0$
And substituting $x=w^2$ in
$1+x^{130}+x^{1988}=1+w^{260}+w^{3976}=1+w^{258+2}+w^{3974+1}=1+w^2+w=0$
Hence
$w,w^2$ are also roots of $1+x^{130}+x^{1988}=0$
Hence, options 'B' and 'A' are correct.