Solve: 3^(-3) × 19^35 × 19^(-32) ÷ 19^3 | Exponent Simplification

Let's start by simplifying the second term in the complete multiplication, meaning - the fraction. We'll simplify it in two stages:

In the first stage we'll use the power law for multiplication between terms with identical bases:

$a^m\cdot a^n=a^{m+n}$

and simplify the fraction's numerator:

$\frac{19^{35}\cdot19^{-32}}{19^3}=\frac{19^{35+(-32)}}{19^3}=\frac{19^{35-32}}{19^3}=\frac{19^3}{19^3}$

Next, we can either remember that dividing any number by itself gives 1, or use the power law for division between terms with identical bases:

$\frac{a^m}{a^n}=a^{m-n}$ to get that:$\frac{19^3}{19^3}=19^{3-3}=19^0=1$

where in the last step we used the fact that raising any number to the power of 0 gives 1, meaning mathematically that:

$X^0=1$

Let's summarize this part, we got that:

$\frac{19^{35}\cdot19^{-32}}{19^3}=1$

Let's now return to the complete expression in the problem and substitute this result in place of the fraction:

$3^{-3}\cdot\frac{19^{35}\cdot19^{-32}}{19^3}=3^{-3}\cdot1=3^{-3}$

In the next stage we'll recall the power law for negative exponents:

$a^{-n}=\frac{1}{a^n}$

and apply this law to the result we got:

$3^{-3}=\frac{1}{3^3}=\frac{1}{27}$

Summarizing all the steps above, we got that:

$3^{-3}\cdot\frac{19^{35}\cdot19^{-32}}{19^3}=3^{-3}=\frac{1}{27}$

Therefore the correct answer is answer A.