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

Welcome, future engineers! Today, we are going to peel back the layers of a problem that, at first glance, looks like a tedious algebraic expansion but is actually a beautiful exercise in geometric transformation.

We are looking at the equation $(x-1{)}^3+8=0$. Many students, upon seeing this, immediately reach for the binomial expansion formula.

They start writing out $x^3-3x^2+3x-1+8=0$, leading to $x^3-3x^2+3x+7=0$. While this is mathematically valid, it is a trap. It leads to a cubic equation that is difficult to factor.

In the world of JEE Advanced, we don't just solve equations; we look for the soul of the equation. Let us solve this with elegance.

The Geometry of the Cube Root

Let us isolate the cubic term. By transposing the $8$, we get:

Now, imagine the complex plane. We know that the cube roots of unity—$1,\omega ,\text{ and }{\omega }^2$—live on the unit circle, spaced perfectly at $120^{\circ }$ intervals. They are the solutions to $Z^3=1$.

Our equation is $Z^3=-8$, where $Z=x-1$. This is not just an equation; it is a transformation. We are taking the unit circle and scaling it by a factor of $\sqrt[3]{-8}$, which is $-2$.

When we multiply the roots of unity by $-2$, we are doing two things simultaneously: we are scaling the radius from $1$ to $2$, and we are rotating the entire system by $180^{\circ }$ because of the negative sign.

So, the three values for $(x-1)$ are simply $-2(1)$, $-2(\omega )$, and $-2({\omega }^2)$. This is the power of complex numbers—we didn't need to perform a single long division or synthetic division. We simply mapped the roots of unity to our new target.

The Final Translation

We have found the values for $(x-1)$, but our journey is not quite over. We need to find $x$.

The equation $x-1=\text{root}$ implies that $x=1+\text{root}$. Geometrically, this is a translation. We take our circle of radius $2$ (which we just created) and shift it to the right by $1$ unit along the real axis.

Let us calculate the final roots explicitly:

1. For the first root: $x-1=-2⟹x=-1$.

2. For the second root: $x-1=-2\omega ⟹x=1-2\omega$.

3. For the third root: $x-1=-2{\omega }^2⟹x=1-2{\omega }^2$.

Why This Matters

Look at the final set of roots: $\{-1,1-2\omega ,1-2{\omega }^2\}$. This matches Option 2 perfectly.

Notice how we never had to deal with the messy cubic polynomial. We treated the equation as a geometric object, manipulated it, and arrived at the solution with absolute clarity.

This is the mindset you need for the JEE Advanced. When you see a high-degree polynomial, pause. Ask yourself: Can I view this as a transformation of a simpler, known object?

If you can, you will save time, reduce errors, and—most importantly—you will start to see the beautiful, interconnected architecture of mathematics. Keep practicing, keep visualizing, and keep pushing boundaries!