How do you simplify ((4^0c^2d^3f)/(2c^-4d^-5))^-3?

Jun 7, 2017

$\frac{8}{{c}^{18} {d}^{24} {f}^{3}}$

Explanation:

So using the multiplication exponent rule it would simplify to

$\frac{{\left(\frac{1}{64}\right)}^{0} {c}^{-} 6 {d}^{-} 9 {f}^{-} 3}{- \frac{1}{8} {c}^{12} {d}^{15}}$

Using Exponent rules again ${\left(\frac{1}{64}\right)}^{0}$ becomes 1

$\frac{{c}^{-} 6 {d}^{-} 9 {f}^{-} 3}{\frac{1}{8} {c}^{12} {d}^{15}}$

Using the subtraction exponent rule (When you subtract exponents due to division) it simplifies to this:

$\frac{1}{\frac{1}{8} {c}^{18} {d}^{24} {f}^{3}}$

Simplifying it further by getting rid of the fraction gives:

$\frac{8}{{c}^{18} {d}^{24} {f}^{3}}$

Sep 5, 2017

$\frac{8}{{c}^{18} {d}^{24} {f}^{3}}$

Explanation:

${\left(\frac{{4}^{0} {c}^{2} {d}^{3} f}{2 {c}^{-} 4 {d}^{-} 5}\right)}^{-} 3$

Recall the law of indices: ${\left(\frac{a}{b}\right)}^{-} m = {\left(\frac{b}{a}\right)}^{m}$

Start by getting rid of the negative indices:

${\left(\frac{{4}^{0} {c}^{2} {d}^{3} f}{2 {c}^{-} 4 {d}^{-} 5}\right)}^{\textcolor{red}{- 3}} = {\left(\frac{2 {c}^{-} 4 {d}^{-} 5}{{4}^{0} {c}^{2} {d}^{3} f}\right)}^{\textcolor{red}{3}} \text{ } \leftarrow$ invert the fraction

Recall the law of indices: ${x}^{- m} = \frac{1}{x} ^ m$

${\left(\frac{2 \textcolor{b l u e}{{c}^{-} 4 {d}^{-} 5}}{{4}^{0} {c}^{2} {d}^{3} f}\right)}^{3} = {\left(\frac{2}{{4}^{0} {c}^{2} {d}^{3} f \times \textcolor{b l u e}{{c}^{4} {d}^{5}}}\right)}^{3}$

Recall the laws: ${x}^{0} = 1 \mathmr{and} {x}^{m} \times {x}^{n} = {x}^{m + n}$

${\left(\frac{2}{{\cancel{{4}^{0}}}^{1} {c}^{2} {d}^{3} f {c}^{4} {d}^{5}}\right)}^{3} = {\left(\frac{2}{{c}^{6} {d}^{8} f}\right)}^{3}$

Recall the law: ${\left({x}^{m}\right)}^{n} = {x}^{m n}$

${\left(\frac{2}{{c}^{6} {d}^{8} f}\right)}^{3} = \frac{8}{{c}^{18} {d}^{24} {f}^{3}} \text{ } \leftarrow$ multiply the indices