How do you simplify (2x^0y^2)^-3*2yx^3 and write it using only positive exponents?

Oct 26, 2017

$\frac{{y}^{-} 5 {x}^{3}}{4}$

Explanation:

${\left(2 {x}^{0} {y}^{2}\right)}^{-} 3 \cdot 2 y {x}^{3}$

laws of indices:

${\left({a}^{m}\right)}^{n} = {a}^{m n}$
${a}^{m} \cdot {a}^{n} = {a}^{m + n}$

${\left({x}^{0}\right)}^{-} 3 = {x}^{0} = 1$

${\left({y}^{2}\right)}^{-} 3 = {y}^{-} 6$

${2}^{-} 3 = \frac{1}{8}$

${\left(2 {x}^{0} {y}^{2}\right)}^{-} 3 = {y}^{-} \frac{6}{8}$

${y}^{-} \frac{6}{8} \cdot 2 y {x}^{3} = 2 \cdot {y}^{-} 5 \cdot {x}^{3} \cdot \frac{1}{8}$

$= \frac{{y}^{-} 5 \cdot {x}^{3}}{4}$

$= \frac{{y}^{-} 5 {x}^{3}}{4}$