Find the 1 st 3 terms of the $$ (1-1 / 3 x)^{-3} $$

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We begin with the binomial expansion for a general exponent:
$$
(1+z)^\alpha=\sum_{n=0}^\infty \binom{\alpha}{n}z^n,
$$
where the binomial coefficient is given by
$$
\binom{\alpha}{n}=\frac{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-n+1)}{n!}.
$$

For the function 
$$
\left(1-\frac{1}{3}x\right)^{-3},
$$
we set
$$
\alpha=-3 \quad \text{and} \quad z=-\frac{1}{3}x.
$$

Now, we compute the first three terms:

1. For $$ n=0 $$:
   $$
   \binom{-3}{0}=1,\quad \left(-\frac{1}{3}x\right)^0=1.
   $$
   Hence, the $$n=0$$ term is
   $$
   1.
   $$

2. For $$ n=1 $$:
   $$
   \binom{-3}{1}=-3,\quad \left(-\frac{1}{3}x\right)^1=-\frac{1}{3}x.
   $$
   Multiplying these, we get
   $$
   -3 \times \left(-\frac{1}{3}x\right)=x.
   $$

3. For $$ n=2 $$:
   $$
   \binom{-3}{2}=\frac{(-3)(-4)}{2}= \frac{12}{2}=6,\quad \left(-\frac{1}{3}x\right)^2=\frac{x^2}{9}.
   $$
   Thus, the $$n=2$$ term is
   $$
   6 \times \frac{x^2}{9}=\frac{6x^2}{9}=\frac{2}{3}x^2.
   $$

Therefore, the first three terms of the expansion are:
$$
1 + x + \frac{2}{3}x^2.
$$
The first three terms of the expansion are $$1 + x + \frac{2}{3}x^2$$.

Find the 1 st 3 terms of the \( (1-1 / 3 x)^{-3} \)

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