JAMIA 2021 — Mathematics PYQ

The given equation is $x^3-2x^2+2x-1=0$.

Notice that $x=1$ is a root because $1^3-2(1{)}^2+2(1)-1=1-2+2-1=0$.

We can factorize $(x-1)$ out:

Step 2: Identify the Roots ($\alpha ,\beta ,\gamma$)

From the factors:

1. $x-1=0⟹x=1$

2. $x^2-x+1=0$

The roots of $x^2-x+1=0$ are:

These are the negative complex cube roots of unity:

(Where $\omega$ and ${\omega }^2$ are the complex cube roots of unity satisfying ${\omega }^3=1$ and ${\omega }^2+\omega +1=0$).

Step 3: Analyze the Power $162$

We need to find ${\alpha }^{162}+{\beta }^{162}+{\gamma }^{162}$.

First, check if $162$ is a multiple of $3$ or $6$:

$162÷3=54$ (exactly)

$162÷6=27$ (exactly)

Now calculate each term:

1. ${\alpha }^{162}=1^{162}=1$

2. ${\beta }^{162}=(-{\omega }^2{)}^{162}=(-1{)}^{162}\cdot ({\omega }^2{)}^{162}=1\cdot {\omega }^{324}$

   Since $324$ is a multiple of $3$ ($3\times 108$), ${\omega }^{324}=1$.

   So, ${\beta }^{162}=1$.

3. ${\gamma }^{162}=(-\omega {)}^{162}=(-1{)}^{162}\cdot {\omega }^{162}=1\cdot {\omega }^{162}$

   Since $162$ is a multiple of $3$ ($3\times 54$), ${\omega }^{162}=1$.

   So, ${\gamma }^{162}=1$.

Step 4: Final Sum

Final Answer:

The value is $3$.