a) x^5. x^6. x^2 б) a^12: a^10: a в)(2x^3)^2 г) (-3x^3y^2)^4

Let's simplify each of the expressions one by one:

a) \( x^5 \cdot x^6 \cdot x^2 \)

Using the property of exponents that states \( a^m \cdot a^n = a^{m+n} \):

\[ x^5 \cdot x^6 \cdot x^2 = x^{5 + 6 + 2} = x^{13} \]

b) \( a^{12} : a^{10} : a \)

First, we interpret the colon ":" as a sequence of divisions. This can be represented as:

\[ \frac{a^{12}}{a^{10}} \cdot \frac{1}{a} \]

Using the property of exponents for division, \( \frac{a^m}{a^n} = a^{m-n} \):

\[ \frac{a^{12}}{a^{10}} = a^{12-10} = a^2 \]

Now, the next part:

\[ \frac{a^2}{a} = a^{2-1} = a^1 = a \]

c) \( (2x^3)^2 \)

Using the power of a product property \( (ab)^n = a^n \cdot b^n \):

\[ (2x^3)^2 = 2^2 \cdot (x^3)^2 = 4 \cdot x^{3 \cdot 2} = 4x^6 \]

d) \( (-3x^3y^2)^4 \)

Again, using the power of a product property:

\[ (-3x^3y^2)^4 = (-3)^4 \cdot (x^3)^4 \cdot (y^2)^4 \]

Calculating each part:

\[ (-3)^4 = 81, \quad (x^3)^4 = x^{3 \cdot 4} = x^{12}, \quad (y^2)^4 = y^{2 \cdot 4} = y^8 \]

Thus, combining these results:

\[ (-3x^3y^2)^4 = 81x^{12}y^8 \]

To summarize:

a) \( x^{13} \)

b) \( a \)

c) \( 4x^6 \)

d) \( 81x^{12}y^8 \)