How do you multiply and simplify ${(\frac{2 x y^{2} \cdot - 2 x^{0} y^{2}}{- 2 x^{- 1} y^{0}})}^{3}$ $(\frac { 2x y ^ { 2} \cdot - 2x ^ { 0} y ^ { 2} } { - 2x ^ { - 1} y ^ { 0} } ) ^ { 3}$ ?

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Answer 1 · 2018-06-26T03:35:32

The answer is $8 x^{6} y^{12}$ $8x^6y^12$ .

${(\frac{2 x y^{2} \cdot (- 2 x^{0} y^{2})}{- 2 x^{- 1} y^{0}})}^{3}$ $((2xy^2*(-2x^0 y^2))/(-2x^-1 y^0))^3$

First, remember that anything raised to the 0th power is equal to $1$ $1$ . So start by canceling out all the terms that are raised to $0$ $0$ .

${(\frac{2 x y^{2} \cdot (- 2 (1) y^{2})}{- 2 x^{- 1} (1)})}^{3}$ $((2xy^2 * (-2(1)y^2))/(-2x^-1(1)))^3$

${(\frac{2 x y^{2} \cdot (- 2 y^{2})}{- 2 x^{- 1}})}^{3}$ $((2xy^2 * (-2y^2))/(-2x^-1))^3$

Multiply the numerator by adding the exponent of like terms. Don't forget the negative sign!

${(\frac{- 4 x y^{4}}{- 2 x^{- 1}})}^{3}$ $((-4xy^4)/(-2x^-1))^3$

${(\frac{2 x y^{4}}{x^{- 1}})}^{3}$ $((2xy^4)/(x^-1))^3$

Remember that anything raised to the $- 1$ $-1$ is basically a reciprocal, so:

$x^{- 1} = \frac{1}{x}$ $x^-1 = 1/x$

Therefore:

${(\frac{2 x y^{4}}{\frac{1}{x}})}^{3}$ $((2xy^4)/(1/x))^3$

which is

${(2 x y^{4} \cdot x)}^{3}$ $(2xy^4 * x)^3$

${(2 x^{2} y^{4})}^{3}$ $(2x^2y^4)^3$

Distribute the exponent

$2^{3} \cdot x^{2 \cdot 3} \cdot y^{4 \cdot 3}$ $2^3 * x^(2 * 3) * y^ (4*3)$

$= 8 x^{6} y^{12}$ $= 8x^6y^12$

How do you multiply and simplify (\frac { 2x y ^ { 2} \cdot - 2x ^ { 0} y ^ { 2} } { - 2x ^ { - 1} y ^ { 0} } ) ^ { 3}(2xy2⋅−2x0y2−2x−1y0)3(\frac { 2x y ^ { 2} \cdot - 2x ^ { 0} y ^ { 2} } { - 2x ^ { - 1} y ^ { 0} } ) ^ { 3}?