Animated Solution for Mathematics - Quadratic Equations: The sum of the cubes of all the roots of the equation $x^4-3x^3-2x^2+3x+1=0$ is ______.

Welcome, future engineers. Today, we are going to dismantle a problem that, at first glance, looks like a terrifying quartic equation. You see $x^4-3x^3-2x^2+3x+1=0$ and your instinct might be to panic.

But I want you to take a deep breath. In the world of JEE Advanced, we don't fight equations with brute force; we fight them with elegance. We look for the hidden soul of the expression.

The Detective Work

Look at the coefficients: $1,-3,-2,3,1$. Do you see it? The coefficients are symmetric, but with a twist—the signs are flipped for the odd powers.

This is a classic reciprocal equation. When you see this, you shouldn't see a wall; you should see a doorway. The symmetry is a gift. It tells us that the variable $x$ and its reciprocal $\frac{1}{x}$ are intimately connected.

If we can group them, we can simplify the entire structure.

The Algebraic Surgery

To unlock this, we perform a bit of algebraic surgery. We divide the entire equation by $x^2$. Why $x^2$? Because it is the middle term, the pivot point of our symmetry.

Dividing by $x^2$ gives us:

Now, let us group the terms that belong together. We pair $x^2$ with $\frac{1}{x^2}$, and we pair $-3x$ with $\frac{3}{x}$. This leaves us with:

This is where the magic happens. We introduce a substitution. Let $t=x-\frac{1}{x}$.

If we square this, we get $t^2=x^2+\frac{1}{x^2}-2$. This means $x^2+\frac{1}{x^2}=t^2+2$.

Suddenly, our terrifying quartic has collapsed into a simple quadratic in $t$:

The Two Worlds

Solving $t(t-3)=0$ gives us two distinct cases: $t=0$ and $t=3$. We have effectively split our four roots into two pairs.

In Case 1, where $t=0$, we have $x-\frac{1}{x}=0$, which leads to $x^2=1$. The roots are $1$ and $-1$. The sum of their cubes is $1^3+(-1{)}^3=0$.

In Case 2, where $t=3$, we have $x-\frac{1}{x}=3$, which rearranges to $x^2-3x-1=0$. Let the roots be $\gamma$ and $\delta$.

Now, listen closely: do not solve for $\gamma$ and $\delta$! If you do, you will be trapped in a maze of square roots. Instead, use the power of Vieta's formulas.

The Masterstroke

We know that for $x^2-3x-1=0$, the sum of the roots $\gamma +\delta =3$ and the product $\gamma \delta =-1$. We need the sum of the cubes, ${\gamma }^3+{\delta }^3$.

We use the identity:

Substitute our values:

Finally, we combine our results. The total sum of the cubes of all four roots is the sum from Case 1 ($0$) plus the sum from Case 2 ($36$).

The result is $36$.

See how we navigated that? We didn't fight the quartic; we transformed it, split it, and conquered it using the beauty of identities. Keep this mindset, and no equation will ever be too daunting for you.