Animated Solution for Mathematics - Complex Numbers: If $\alpha$ satisfies the equation $x^2+x+1=0$ and $(1+\alpha {)}^7=A+B\alpha +C{\alpha }^2$, $A,B,C\ge 0$, then $5(3A-2B-C)$ is equal to

Welcome, fellow explorer of the mathematical universe. Today, we are going to dissect a problem that, at first glance, might look like a tedious exercise in binomial expansion.

But as we peel back the layers, we will find that it is actually a beautiful dance with the complex cube roots of unity. If you have ever felt intimidated by the sight of $\omega$, take a deep breath. We are going to master it together.

The Hidden Identity

We begin with the quadratic equation $x^2+x+1=0$. In the world of JEE Advanced, this is a signal, a beacon.

Whenever you see this, your mind should immediately jump to the complex cube roots of unity. If we multiply both sides by $(x-1)$, we get:

This implies that the roots of our equation are the non-real cube roots of unity, which we denote as $\omega$ and ${\omega }^2$. Let us set $\alpha =\omega$. This is our anchor.

Recall the fundamental identity that governs these roots: $1+\omega +{\omega }^2=0$. This is the secret key to the entire problem.

It tells us that the sum of the roots is $-1$, and more importantly, it allows us to express any power of $\omega$ in terms of lower powers. Specifically, we can rearrange this to find that $1+\omega =-{\omega }^2$. This single substitution is going to save us from pages of unnecessary algebra.

Taming the Expression

The problem asks us to evaluate $(1+\alpha {)}^7$. With our identity $1+\alpha =-{\omega }^2$ in hand, the expression transforms instantly.

Instead of dealing with a binomial raised to the seventh power, we are now looking at $(-{\omega }^2{)}^7$. This is the power of mathematical intuition—we have replaced a complex binomial with a single, manageable term.

Now, let us handle the exponent. We have $(-{\omega }^2{)}^7$. Since the exponent $7$ is odd, the negative sign persists: $-({\omega }^2{)}^7$. Applying the power of a power rule, we get $-{\omega }^{14}$.

The Reduction

Here is where the magic happens. We know that ${\omega }^3=1$. This means that any power of $\omega$ can be reduced by dividing the exponent by $3$ and looking at the remainder.

For ${\omega }^{14}$, we can write this as ${\omega }^{12}\cdot {\omega }^2$. Since ${\omega }^{12}=({\omega }^3{)}^4=1^4=1$, the entire term ${\omega }^{14}$ simplifies beautifully to just ${\omega }^2$.

So, our expression $(1+\alpha {)}^7$ has collapsed from a daunting binomial to simply $-{\omega }^2$. But we are not done yet! We need to compare this to the form $A+B\alpha +C{\alpha }^2$.

Let us use our identity $1+\omega +{\omega }^2=0$ one last time. We know that $-{\omega }^2=1+\omega$. Since $\alpha =\omega$, this is exactly $1+\alpha$.

The Final Comparison

We have arrived at the finish line. We have shown that $(1+\alpha {)}^7=1+\alpha$. To match this with the required form $A+B\alpha +C{\alpha }^2$, we write it as $1+1\alpha +0{\alpha }^2$.

By comparing the coefficients, we see clearly that $A=1$, $B=1$, and $C=0$.

Finally, we calculate the value of $5(3A-2B-C)$. Substituting our values, we get:

Look at what we have achieved. We took a problem that could have been a nightmare of binomial coefficients and, by using the properties of $\omega$, reduced it to a simple linear comparison.

This is the essence of JEE Advanced mathematics: it is not about brute force; it is about finding the elegant path. Keep this mindset, and you will find that even the most complex problems have a simple, beautiful core waiting to be discovered. You have done well today.