Animated Solution for Mathematics - Binomial Theorem: The sum of the co-efficients of all even degree terms in $x$ in the expansion of $(x+\sqrt{x^3-1}{)}^6+(x-\sqrt{x^3-1}{)}^6,(x>1)$ is equal to :

Imagine you are staring at the expression $(x+\sqrt{x^3-1}{)}^6+(x-\sqrt{x^3-1}{)}^6$. At first glance, it looks like a nightmare. That square root term, $\sqrt{x^3-1}$, seems designed to make your life difficult.

But here is the secret of JEE Advanced: often, the most intimidating problems are just standard identities wearing a disguise. This expression is a perfect example of the binomial identity $(a+b{)}^n+(a-b{)}^n$.

When you see this structure, your first instinct should be to celebrate, because it is a gift that simplifies everything.

Let us map our variables: $a=x$, $b=\sqrt{x^3-1}$, and $n=6$. When we expand $(a+b{)}^6$ and $(a-b{)}^6$ and add them together, something magical happens.

All the terms with odd powers of $b$—which contain the square root—cancel out completely. We are left with the following expression:

This is our roadmap. We do not need to worry about the square root for long, because as we calculate the terms, the powers of $b$ will be even, effectively squaring away that root.

Let's break this down step-by-step. First, the term $T_0$ is:

Next, we evaluate $T_2$:

Now, we calculate $T_4$:

Finally, we calculate $T_6$:

To find the sum of the coefficients of the even-degree terms, we extract the relevant values from our expansions:

From $T_0$: $1$ (coefficient of $x^6$)

From $T_2$: $-15$ (coefficient of $x^4$)

From $T_4$: $15$ (coefficient of $x^8$) and $15$ (coefficient of $x^2$)

From $T_6$: $-3$ (coefficient of $x^6$) and $-1$ (constant term $x^0$)

Summing these coefficients inside the bracket, we get:

Remember our identity: the total sum is $2\times 12$.

The final result is $24$. You have navigated the complexity, identified the pattern, and arrived at the elegant solution.