1) $$ 2 k^{3} $$ k) $$ (2 k)^{-3} $$ m) $$ (-2 x)^{-3} $$ i) $$ 4 x^{-1} y^{-2} $$ 0) $$ -5 x^{-3} y y^{-2} $$ P) $$ 2^{-2} x^{6} y^{-3} $$

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1) $$ 2 k^{3} $$

The expression is already in simplest form.

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k) $$ (2k)^{-3} $$

We apply the rule for negative exponents:
$$
(2k)^{-3} = \frac{1}{(2k)^3}
$$
 Now, evaluate the cube:
$$
(2k)^3 = 2^3 \cdot k^3 = 8k^3
$$
 Thus, the simplified form is:
$$
(2k)^{-3} = \frac{1}{8k^3}
$$

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m) $$ (-2x)^{-3} $$

Similarly, use the negative exponent rule:
$$
(-2x)^{-3} = \frac{1}{(-2x)^3}
$$
 Compute the cube:
$$
(-2x)^3 = (-2)^3 \cdot x^3 = -8x^3
$$
 So we have:
$$
(-2x)^{-3} = \frac{1}{-8x^3} = -\frac{1}{8x^3}
$$

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i) $$ 4x^{-1}y^{-2} $$

Convert the negative exponents into fractions:
$$
x^{-1} = \frac{1}{x} \quad \text{and} \quad y^{-2} = \frac{1}{y^2}
$$
 Thus,
$$
4x^{-1}y^{-2} = 4 \cdot \frac{1}{x} \cdot \frac{1}{y^2} = \frac{4}{xy^2}
$$

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0) $$ -5x^{-3}yy^{-2} $$

First, combine the $$ y $$ terms. Recall that:
$$
y \cdot y^{-2} = y^{1-2} = y^{-1}
$$
 So the expression becomes:
$$
-5x^{-3}y^{-1}
$$
 Rewrite the negative exponents as fractions:
$$
x^{-3} = \frac{1}{x^3} \quad \text{and} \quad y^{-1} = \frac{1}{y}
$$
 Thus,
$$
-5x^{-3}y^{-1} = -5 \cdot \frac{1}{x^3} \cdot \frac{1}{y} = -\frac{5}{x^3y}
$$

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P) $$ 2^{-2}x^{6}y^{-3} $$

First, simplify the constant:
$$
2^{-2} = \frac{1}{2^2} = \frac{1}{4}
$$
 Also, convert $$ y^{-3} $$ to a fraction:
$$
y^{-3} = \frac{1}{y^3}
$$
 Thus, the expression becomes:
$$
2^{-2}x^{6}y^{-3} = \frac{1}{4} \cdot x^6 \cdot \frac{1}{y^3} = \frac{x^6}{4y^3}
$$
1) $$ 2k^{3} $$ - Already in simplest form. k) $$ \frac{1}{8k^{3}} $$ - Simplified from $$ (2k)^{-3} $$. m) $$ -\frac{1}{8x^{3}} $$ - Simplified from $$ (-2x)^{-3} $$. i) $$ \frac{4}{xy^{2}} $$ - Simplified from $$ 4x^{-1}y^{-2} $$. 0) $$ -\frac{5}{x^{3}y} $$ - Simplified from $$ -5x^{-3}yy^{-2} $$. P) $$ \frac{x^{6}}{4y^{3}} $$ - Simplified from $$ 2^{-2}x^{6}y^{-3} $$.

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