If the cube roots of unity are 1, ω, ω², then the roots of the equation (x-1)³ + 8 = 0 are (A) -1, 1

Q: Why can't I just expand $(x-1)^3 + 8 = 0$ and solve the cubic?

You absolutely can, but it is a path filled with unnecessary algebraic labor! Expanding $(x-1)^3$ gives $x^3 - 3x^2 + 3x - 1 + 8 = x^3 - 3x^2 + 3x + 7 = 0$. Solving this cubic requires finding one real root by inspection and then performing polynomial division. Using the properties of the roots of unity is the 'JEE way'—it is faster, more elegant, and reveals the beautiful geometric structure of the roots.

Q: Why do we get three roots? I thought the answer was just $-1$.

That is a classic misconception! The Fundamental Theorem of Algebra tells us that a polynomial of degree $n$ must have exactly $n$ roots in the complex number system. Since our equation is a cubic (degree 3), we must find three roots. The real root $x = -1$ is just the tip of the iceberg; the other two are complex numbers that satisfy the equation.

Q: What exactly is $\omega$ and why does it appear here?

The symbol $\omega$ represents the complex cube root of unity, defined as $\omega = e^{i2\pi/3} = \frac{-1 + i\sqrt{3}}{2}$. It is the engine of symmetry in complex numbers. When we solve $z^3 = 1$, we don't just get $1$; we get $1, \omega, \text{ and } \omega^2$. These roots are perfectly spaced at $120^\circ$ intervals on the unit circle. Whenever you see a cubic equation, think of $\omega$ as your key to unlocking the hidden complex solutions.

Question

If the cube roots of unity are $1, \omega, \omega^2$, then the roots of the equation $(x-1)^3 + 8 = 0$ are

Options:

(A) $-1, 1 + 2\omega, 1 + 2\omega^2$

(B) $-1, 1 - 2\omega, 1 - 2\omega^2$

(C) $-1, -1, -1$

(D) None of these

Sigma Percentile · Accepted Answer

### The Elegance of the Complex Plane

Welcome, future IITians! Today, we are going to dismantle a problem that often trips up students in the JEE Advanced examination. The equation is $(x-1)^3 + 8 = 0$.

At first glance, it looks like a standard cubic equation. Many students will immediately reach for the expansion formula, turning this into a messy polynomial. But I want you to pause.

In JEE Advanced, the most powerful tool you have is not your speed of calculation, but your ability to see the structure of the problem.

### Phase 1: The Shift

Look at the equation $(x-1)^3 + 8 = 0$. Do you see the structure? We have a perfect cube plus a constant.

Let us isolate the cubic term: 
$$(x-1)^3 = -8$$

Now, instead of expanding, let us perform a mental shift. Let $Y = x-1$. Our equation becomes $Y^3 = -8$.

This is much cleaner, isn't it? We know that $-8$ is simply $(-2)^3$. So, we have $Y^3 = (-2)^3$. This is the moment where the magic of complex numbers begins.

### Phase 2: The Geometry of Roots

When we solve $Y^3 = (-2)^3$, we are essentially looking for the cube roots of $-8$. If we divide by $(-2)^3$, we get:

$$\left( \frac{Y}{-2} ight)^3 = 1$$

This means the values of $\frac{Y}{-2}$ are the cube roots of unity: $1, \omega, 	ext{ and } \omega^2$. Therefore, our three values for $Y$ are $-2(1), -2(\omega), 	ext{ and } -2(\omega^2)$.

Imagine standing at the origin of the complex plane. The roots of unity $1, \omega, \omega^2$ form an equilateral triangle inscribed in a unit circle.

By multiplying these by $-2$, we are simply scaling this triangle by a factor of $2$ and reflecting it across the origin. We have effectively transformed our problem into a geometric rotation and scaling.

### Phase 3: The Final Calculation

Now, we simply substitute back $Y = x-1$. We have three cases:

1. $x - 1 = -2 \implies x = -1$

2. $x - 1 = -2\omega \implies x = 1 - 2\omega$

3. $x - 1 = -2\omega^2 \implies x = 1 - 2\omega^2$

These are our three roots! Notice how the real root, $x = -1$, sits on the real axis, while the other two are complex conjugates of each other.

If you were to plot these on the complex plane, you would see a perfect equilateral triangle centered at the point $(1, 0)$ with a radius of $2$.

### Conclusion

This problem is a masterclass in why we study complex numbers. By avoiding the brute-force expansion, we not only saved time but also gained a deeper understanding of the symmetry inherent in cubic equations.

The roots are **$-1, 1 - 2\omega, 	ext{ and } 1 - 2\omega^2$**.

Keep this perspective in your toolkit—whenever you see a cubic equation, look for the shift, look for the roots of unity, and look for the geometry. You are not just solving for $x$; you are mapping the structure of the universe.

Animated Solution for Mathematics - Complex Numbers: If the cube roots of unity are $1, ω, ω^{2}$ , then the roots of the equation $(x - 1)^{3} + 8 = 0$ are

Select Answer:

Visualized Solution

Analyze the Equation $(x - 1)^{3} + 8 = 0$

Isolate the Cubic Term

Express as a Perfect Cube

Take the Cube Root

Apply Cube Roots of Unity

Solve for the First Root

Solve for the Second Root

Solve for the Third Root

Conclusion and Geometric Interpretation

The Sigma Insight: Cube Roots and nth Roots of Unity

The Elegance of the Complex Plane

Phase 1

Phase 2

Phase 3

Conclusion

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Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the cube roots of unity are 1,ω,ω2, then the roots of the equation (x−1)3+8=0 are