Question

1. \\(\frac{3x^{3}y^{-1}z^{-1}}{x^{-4}y^{0}z^{0}}\\)

Explanation:

Step1: Simplify zero exponents

Any non-zero term to the 0 power is 1, so $y^0=1$, $z^0=1$. The denominator becomes $x^{-4} \cdot 1 \cdot 1 = x^{-4}$.
The expression is now $\frac{3x^3 y^{-1} z^{-1}}{x^{-4}}$

Step2: Simplify $x$ terms

Use exponent rule $\frac{x^a}{x^b}=x^{a-b}$.
$x^{3 - (-4)} = x^{3+4}=x^7$
The expression is now $3x^7 y^{-1} z^{-1}$

Step3: Rewrite negative exponents

Use rule $x^{-a}=\frac{1}{x^a}$.
$y^{-1}=\frac{1}{y}$, $z^{-1}=\frac{1}{z}$

Answer:

$\frac{3x^7}{yz}$