The sum of 162th power of the roots of the equation x3 - 2x2 + 2x - 1 = 0 is \underline\hspace2cm .

Quadratic Equation and Inequalities PYQ — JEE Main 2021 · 26 Feb · Shift 1 · Q39 with Solution

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Correct answer: 3

We are given: $x^3-2x^2+2x-1=0$

Group terms: $x^3-2x^2+2x-1 = x^2(x-2)+1(2x-1)$ A better way is to check simple roots.

Substitute $x=1$ : $1-2+2-1=0$ So $(x-1)$ is a factor.

Now divide: $x^3-2x^2+2x-1=(x-1)(x^2-x+1)$

Thus the roots are:

\qquad x=\frac{1+i\sqrt3}{2}, \qquad x=\frac{1-i\sqrt3}{2}$$ The complex roots are: $$\omega=\frac{1+i\sqrt3}{2}=\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}$$ $$\bar\omega=\frac{1-i\sqrt3}{2}=\cos\frac{\pi}{3}-i\sin\frac{\pi}{3}$$ So they are $e^{i\pi/3}$ and $e^{-i\pi/3}$. --- 2. **Compute the 162th powers** We need: $$1^{162}+\omega^{162}+\bar\omega^{162}$$ Clearly, $$1^{162}=1$$ Now, $$\omega^{162}=\left(e^{i\pi/3}\right)^{162}=e^{i54\pi}$$ Since $54\pi=27\cdot 2\pi$, $$e^{i54\pi}=1$$ Similarly, $$\bar\omega^{162}=\left(e^{-i\pi/3}\right)^{162}=e^{-i54\pi}=1$$ --- 3. **Add them** Therefore, $$1+1+1=3$$ So the sum of the $162$th powers of the roots is: $$\boxed{3}$$ --- 4. **Comparison with stored correct answer** Stored correct answer = $3$. Our derived answer is also $3$, so it agrees.

View written solutionFree

Correct answer: 3

We are given: $x^3-2x^2+2x-1=0$

Group terms: $x^3-2x^2+2x-1 = x^2(x-2)+1(2x-1)$ A better way is to check simple roots.

Substitute $x=1$ : $1-2+2-1=0$ So $(x-1)$ is a factor.

Now divide: $x^3-2x^2+2x-1=(x-1)(x^2-x+1)$

Thus the roots are:

\qquad x=\frac{1+i\sqrt3}{2}, \qquad x=\frac{1-i\sqrt3}{2}$$ The complex roots are: $$\omega=\frac{1+i\sqrt3}{2}=\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}$$ $$\bar\omega=\frac{1-i\sqrt3}{2}=\cos\frac{\pi}{3}-i\sin\frac{\pi}{3}$$ So they are $e^{i\pi/3}$ and $e^{-i\pi/3}$. --- 2. **Compute the 162th powers** We need: $$1^{162}+\omega^{162}+\bar\omega^{162}$$ Clearly, $$1^{162}=1$$ Now, $$\omega^{162}=\left(e^{i\pi/3}\right)^{162}=e^{i54\pi}$$ Since $54\pi=27\cdot 2\pi$, $$e^{i54\pi}=1$$ Similarly, $$\bar\omega^{162}=\left(e^{-i\pi/3}\right)^{162}=e^{-i54\pi}=1$$ --- 3. **Add them** Therefore, $$1+1+1=3$$ So the sum of the $162$th powers of the roots is: $$\boxed{3}$$ --- 4. **Comparison with stored correct answer** Stored correct answer = $3$. Our derived answer is also $3$, so it agrees.