Find the term independent of x in (x + 1/x)^8

Question

Find the term independent of $x$ in the expansion of $\left(x + \dfrac{1}{x}\right)^8$.

Solution — Step by Step

Why This Works

In a binomial expansion, each term has a specific power of $x$ determined by how we distribute the $x$ and $1/x$ factors. When $r$ copies of $1/x$ are chosen (contributing $x^{-r}$) and $(8-r)$ copies of $x$ (contributing $x^{8-r}$), the net power is $x^{8-r-r} = x^{8-2r}$.

For this to be “independent of $x$” (i.e., a pure constant), we need $8 - 2r = 0$, which pins $r = 4$ uniquely. This is the term where exactly half the factors are $x$ and half are $1/x$, so they cancel perfectly.

This technique — set the exponent of $x$ to zero and solve for $r$ — works for any “term independent of $x$” problem in binomial expansions.

Alternative Method

If the problem were $\left(x^2 + \dfrac{1}{x}\right)^9$, the general term would be $\binom{9}{r} (x^2)^{9-r} \cdot x^{-r} = \binom{9}{r} x^{18-3r}$. Setting $18 - 3r = 0$ gives $r = 6$. The method scales directly.

The key formula to remember: general term $T_{r+1} = \binom{n}{r} a^{n-r} b^r$, then solve for $r$ from the condition on the power of $x$.

Common Mistake