# How do you simplify (((-m^4)(p^4)(q^4))^-1 ) /((m)(q)(p^3) x (m^3)(p^-3)(q^-1)?

Jun 25, 2017

color(blue)(1/(-m^8p^4q^4x)

#### Explanation:

${\left(\left(- {m}^{4}\right) \left({p}^{4}\right) \left({q}^{4}\right)\right)}^{-} \frac{1}{\left(m\right) \left(q\right) \left({p}^{3}\right) x \left({m}^{3}\right) \left({p}^{-} 3\right) \left({q}^{-} 1\right)}$

$\therefore = \frac{\frac{1}{- {m}^{4} \cdot {p}^{4} \cdot {q}^{4}}}{m \cdot q \cdot {p}^{3} \cdot x \cdot {m}^{3} \cdot {p}^{-} 3 \cdot {q}^{-} 1}$

$\therefore = \frac{\frac{1}{- {m}^{4} \cdot {p}^{4} \cdot {q}^{4}}}{{m}^{1 + 3} \cdot {q}^{1 - 1} \cdot {p}^{3 - 3} \cdot x}$

$\therefore = \frac{\frac{1}{- {m}^{4} \cdot {p}^{4} \cdot {q}^{4}}}{{m}^{4} \cdot {q}^{0} \cdot {p}^{0} \cdot x}$

$\therefore = \frac{\frac{1}{- {m}^{4} \cdot {p}^{4} \cdot {q}^{4}}}{{m}^{4} \cdot 1 \cdot 1 \cdot x}$

$\therefore = \frac{\frac{1}{- {m}^{4} \cdot {p}^{4} \cdot {q}^{4}}}{{m}^{4} \cdot x}$

$\therefore = \frac{1}{- {m}^{4} \cdot {p}^{4} \cdot {q}^{4}} \times \frac{1}{{m}^{4} \cdot x}$

$\therefore = \frac{1}{- {m}^{4 + 4} \cdot {p}^{4} \cdot {q}^{4} \cdot x}$

:.color(blue)(=1/(-m^8*p^4*q^4*x)

Jun 25, 2017

$- \frac{1}{{m}^{8} {p}^{4} {q}^{4}}$

#### Explanation:

I suspect the $x$ was meant to be $\times$?

Simplify where possible first in the numerator and denominator separately.

${\left(\left(- {m}^{4}\right) \left({p}^{4}\right) \left({q}^{4}\right)\right)}^{-} \frac{1}{\left(m\right) \left(q\right) \left({p}^{3}\right) \times \left({m}^{3}\right) \left({p}^{-} 3\right) \left({q}^{-} 1\right)}$

=((-m^-4)(p^-4)(q^-4))/((m^4)(q^0)(p^0))" "(larr"multiply the indices by -1")/(larr "add the indices of like bases")

${x}^{0} = 1 \mathmr{and} {x}^{-} m = \frac{1}{x} ^ m$

$- \frac{1}{{m}^{4} {m}^{4} {p}^{4} {q}^{4}}$

$- \frac{1}{{m}^{8} {p}^{4} {q}^{4}}$