Q. \[ \frac{3 x^3 y^{-1} z^{-1}}{x^{-4} y^0 z^0} \]

Answer

\[
\frac{3x^{3}y^{-1}z^{-1}}{x^{-4}y^{0}z^{0}}
=3x^{3-(-4)}y^{-1-0}z^{-1-0}
=3x^{7}y^{-1}z^{-1}
=\frac{3x^{7}}{yz}.
\]

Write the original expression clearly:

\( \displaystyle \frac{3x^{3}y^{-1}z^{-1}}{x^{-4}y^{0}z^{0}} \)
Use the exponent rule for quotients: for any nonzero base a, \( \displaystyle \frac{a^{m}}{a^{n}} = a^{m-n} \). Apply this rule separately to each variable (and note that constants divide normally):

x-exponent: \(3 – (-4) = 3 + 4 = 7\).

y-exponent: \(-1 – 0 = -1\).

z-exponent: \(-1 – 0 = -1\).

So the expression becomes

\( \displaystyle 3x^{7}y^{-1}z^{-1} \).
Convert negative exponents to positive by using \(a^{-1} = \frac{1}{a}\):

\( \displaystyle y^{-1} = \frac{1}{y} \) and \( \displaystyle z^{-1} = \frac{1}{z} \).

Therefore

\( \displaystyle 3x^{7}y^{-1}z^{-1} = \frac{3x^{7}}{yz} \).
Final simplified result:

\( \displaystyle \frac{3x^{7}}{yz} \).

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Simplify by subtracting exponents: x^{3-(-4)}=x^{7}, y^{-1-0}=y^{-1}, z^{-1-0}=z^{-1}. Result: 3x^{7}y^{-1}z^{-1}, equivalently 3x^{7}/(yz).

Use a^{m}/a^{n}=a^{m-n}. Here divide x^{3} by x^{-4} to get x^{3-(-4)}=x^{7}.

a^{-n}=1/a^{n}. Negative exponents indicate reciprocals; move factor to the denominator to remove the negative sign.

Any nonzero base to the zero power equals 1, so y^{0}=1 and z^{0}=1. They multiply by 1 and do not change the expression.

The denominator has no numeric factor (implicitly 1), so 3/1 remains 3. Numerical coefficients divide normally like variables.

Move negative-exponent factors to the denominator: 3x^{7}y^{-1}z^{-1}=3x^{7}/(yz).

Practice exponent rules for fluency.
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