Animated Solution for Mathematics - Complex Numbers: If $f(x)$ and $g(x)$ are two polynomials such that the polynomial $P(x)=f(x^3)+xg(x^3)$ is divisible by $x^2+x+1$, then $P(1)$ is equal to _____.

Welcome, fellow traveler on the JEE journey. Today, we are going to dissect a problem that might look like a daunting algebraic puzzle, but beneath the surface, it is a beautiful dance between polynomials and the complex roots of unity.

When you first see $P(x)=f(x^3)+xg(x^3)$ being divisible by $x^2+x+1$, it is natural to feel a bit overwhelmed. But let's take a breath and look at the structure; we are not just solving for $x$, but uncovering the hidden symmetry of the problem.

The Target

Before we dive into the complex math, let's look at our destination. We need to find $P(1)$.

Looking at the definition $P(x)=f(x^3)+xg(x^3)$, if we simply substitute $x=1$, we get:

This is our target. If we can find the values of $f(1)$ and $g(1)$, the problem is essentially solved. Keep this goal in mind; it is the lighthouse guiding us through the fog.

The Master Key

The divisor $x^2+x+1$ is not just any quadratic. It is a fundamental building block in the world of complex numbers.

If you set $x^2+x+1=0$, you get the complex cube roots of unity, $\omega$ and ${\omega }^2$. These roots are special because they satisfy the property ${\omega }^3=1$.

Geometrically, they sit on the unit circle in the complex plane, perfectly spaced. This is our master key. The Factor Theorem tells us that if $P(x)$ is divisible by $x^2+x+1$, then $P(\omega )=0$ and $P({\omega }^2)=0$.

The Factor Theorem in Action

Let's apply the Factor Theorem. Substituting $x=\omega$ into $P(x)$, we get:

Because ${\omega }^3=1$, this equation collapses into $f(1)+\omega g(1)=0$. This is our first equation.

Now, let's do the same for the second root, ${\omega }^2$. Substituting $x={\omega }^2$ into $P(x)$, we get:

Since $({\omega }^2{)}^3={\omega }^6=({\omega }^3{)}^2=1^2=1$, this equation becomes $f(1)+{\omega }^{2}g(1)=0$. This is our second equation.

The Resolution

We now have a system of two linear equations: 1. $f(1)+\omega g(1)=0$ 2. $f(1)+{\omega }^{2}g(1)=0$

Subtracting the second from the first, we get $(\omega -{\omega }^2)g(1)=0$. Since $\omega eq \omega^2$, the term $(\omega -{\omega }^2)$ is non-zero, which forces $g(1)=0$.

Substituting $g(1)=0$ back into the first equation, we immediately find that $f(1)=0$. Finally, our target is:

The elegance of this result is why we love mathematics. We started with a complex-looking polynomial and, through the symmetry of roots, arrived at a clean, zero result. You have mastered the logic; now go forth and apply this to your next challenge!