Animated Solution for Mathematics - Binomial Theorem: In the expansion of $(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}{)}^n,n\in \mathbb{N}$, if the ratio of $15^{th}$ term from the beginning to the $15^{th}$ term from the end is $\frac{1}{6}$, then the value of ${}^n{C}_3$ is:

Welcome, student. Today, we are not just solving a problem; we are uncovering the hidden symmetry within a binomial expansion. When you look at an expression like $(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}{)}^n$, it is easy to feel overwhelmed by the radicals.

But remember, in the world of JEE Advanced, radicals are just exponents in disguise. Let us peel back the layers together.

First, let us transform our expression into a language that algebra loves. We rewrite $\sqrt[3]{2}$ as $2^{1/3}$ and $\frac{1}{\sqrt[3]{3}}$ as $3^{-1/3}$. Now, our expression is $(2^{1/3}+3^{-1/3}{)}^n$.

By setting $a=2^{1/3}$ and $b=3^{-1/3}$, we have simplified our mental model. We are now working with a standard binomial form $(a+b{)}^n$. The general term formula is:

For the $15^{th}$ term from the beginning, we set $r=14$. Thus, our term is:

Now, here is where many students stumble. The problem asks for the $15^{th}$ term from the end. Do we need to expand the whole thing backwards? Absolutely not!

We invoke the beautiful property of binomial symmetry: the $k^{th}$ term from the end is the $(n-k+2{)}^{th}$ term from the beginning. For $k=15$, this becomes the $(n-15+2{)}^{th}$ term, or simply the $(n-13{)}^{th}$ term.

Using our general formula, this gives us:

Because ${}^n{C}_{n-14}$ is identical to ${}^n{C}_{14}$, we have a perfect setup for cancellation.

We are given that the ratio $\frac{T_{15}}{T_{15}^{\prime }}=\frac{1}{6}$. When we place our expressions into this ratio, the binomial coefficients ${}^n{C}_{14}$ vanish into thin air.

This is the moment of clarity! We are left with the ratio of the powers of 2 and 3. By subtracting exponents, we simplify the expression to:

Notice the beauty here: both bases, 2 and 3, end up with the same exponent, $\frac{n-28}{3}$. Because the exponents are identical, we can combine the bases:

We have arrived at $6^{\frac{n-28}{3}}=6^{-1}$.

With the bases equal, we equate the exponents:

A quick multiplication gives $n-28=-3$, leading us to $n=25$.

Finally, we calculate ${}^{25}{C}_3$. Using the formula:

There you have it. What seemed like a terrifying wall of radicals was actually a perfectly balanced equation waiting for you to simplify it. Keep this mindset—look for the symmetry, simplify the bases, and trust the process. You are ready for the next challenge.