# How do you multiply a^(-4)*(a^4b^3)^3?

Apr 6, 2015

${a}^{-} 4 \left({a}^{12} {b}^{9}\right)$

Since ${a}^{-} 4 = \frac{1}{a} ^ 4$
$\frac{{a}^{12} {b}^{9}}{a} ^ 4$

When you divide exponents with the same base, you can subtract them:
${a}^{8} {b}^{9}$

Apr 6, 2015

The result is ${a}^{8} {b}^{9}$.

${a}^{- 4} \cdot {\left({a}^{4} {b}^{3}\right)}^{3}$

Take care of the parentheses first.
${\left({a}^{4} {b}^{3}\right)}^{3}$

Multiply the exponent outside the parentheses by each of the exponents inside the parentheses.

$\left({a}^{4 \times 3} {b}^{3 \times 3}\right) = \left({a}^{12} {b}^{9}\right)$

Now multiply the bases a and b and add the exponents on like bases.

${a}^{- 4} \cdot \left({a}^{12} {b}^{9}\right)$ =

${a}^{- 4 + 12} \left({b}^{9}\right)$ =

${a}^{8} {b}^{9}$

Another way to multiply these terms is to use the inverse of ${a}^{- 4}$.

${a}^{- 4} = \frac{1}{{a}^{4}}$

$\frac{{a}^{12} {b}^{9}}{{a}^{4}}$

Subtract the exponent on base a in the denominator from the exponent on base a in the numerator.

${a}^{12 - 4} {b}^{9}$ = ${a}^{8} {b}^{9}$