Animated Solution for Mathematics - Binomial Theorem: If $x$ is so small that $x^3$ and higher powers of $x$ may be neglected, then $\frac{(1+x{)}^{3/2}-(1+\frac{1}{2}x{)}^3}{(1-x{)}^{1/2}}$ may be approximated as

When you look at an expression like

most students see a wall of complexity. They see fractional powers and division, and they panic. But you see a playground—you see the Binomial Theorem.

The problem provides a golden key: $x$ is so small that $x^3$ and higher powers are negligible. This is not just a constraint; it is a permission slip to simplify the universe.

Why do we love the Binomial Theorem? Because it allows us to turn complex, non-linear functions into simple, manageable polynomials.

The formula is our most reliable tool:

Notice that we stop at the $x^2$ term. Because the problem states that $x^3$ and beyond are effectively zero, calculating terms that will eventually be discarded is a luxury you cannot afford. We are looking for the 'signal' in the noise, and the signal here lives in the $x^2$ term.

Let us dissect the numerator. We have two distinct parts: $(1+x{)}^{3/2}$ and $(1+\frac{1}{2}x{)}^3$.

For the first term, $(1+x{)}^{3/2}$, our index $n$ is $\frac{3}{2}$. Plugging this into our expansion:

Now, for the second term, $(1+\frac{1}{2}x{)}^3$, we treat $\frac{x}{2}$ as a single unit with $n=3$:

Now, we bring them together. We subtract the second expansion from the first:

The constant $1$ vanishes, and the linear term $\frac{3}{2}x$ vanishes. It is a moment of pure mathematical elegance. We are left with:

We have $-\frac{3}{8}{x}^2$ in the numerator and $(1-x{)}^{1/2}$ in the denominator. We bring the denominator up as $(1-x{)}^{-1/2}$ and apply the Binomial Theorem:

Now, multiply this by our numerator:

Recall our constraint: we must neglect $x^3$. Discarding the higher-order term, we arrive at our final, beautiful answer: