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

Solution

Maths sir · 4 steps

Spot x = 1 as a root first, then the other two roots sit on the unit circle at angle ±π/3.

1. Factor the cubic. The coefficients add to 1 - 2 + 2 - 1 = 0, so x = 1 is a root. Dividing out (x - 1) gives x³ - 2x² + 2x - 1 = (x - 1)(x² - x + 1).
2. Polar form. The quadratic gives x = (1 ± i√3)/2 = cos π/3 ± i sin π/3. Both have modulus 1 and argument ±π/3.

   (cos π/3 ± i sin π/3)¹⁶² = cos 54π ± i sin 54π = 1

3. Apply De Moivre. Raising to the 162nd power multiplies the angle by 162, giving 54π. That is 27 full turns of 2π, so each complex root becomes exactly 1. The real root also gives 1¹⁶² = 1.
4. So. The sum is 1 + 1 + 1 = 3.

The common slip: 0. Treating ω¹⁶² + ω³²⁴ like ω + ω² = -1 gives a total of 0. But 162 is a multiple of 3, so each of those powers is 1, not a non-real cube root of unity.

The idea to keep. Write unit-circle roots as cos θ + i sin θ. A power n turns the angle into nθ. If nθ is a multiple of 2π, that power is exactly 1.

Answer from the NTA final answer key. Solution by PrepWiser.