Simplify powers of products worksheet with 14 algebraic expressions, ideal for math practice.

Explanation:
We are asked to simplify expressions that involve powers of products. The key rules we’ll use are:

1. Power of a product: $(ab)^n = a^n b^n$
2. Power of a power: $(a^m)^n = a^{m \cdot n}$
3. Product of same bases: $a^m \cdot a^n = a^{m+n}$
4. Only positive exponents in final answer.

Let’s go one by one.

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1) $(6w^6)^3$
Break it up: $6^3 \cdot (w^6)^3 = 216 \cdot w^{6 \cdot 3} = 216w^{18}$

2) $(2h \cdot 3h^3)^2$
First combine inside: $2 \cdot 3 = 6$, and $h \cdot h^3 = h^{1+3} = h^4$, so inside is $6h^4$.
Now square: $(6h^4)^2 = 6^2 \cdot (h^4)^2 = 36 \cdot h^{8} = 36h^8$

3) $(4g \cdot 3g^3 \cdot g^2)^2$
Multiply coefficients: $4 \cdot 3 = 12$
Multiply powers of $g$: $g^1 \cdot g^3 \cdot g^2 = g^{1+3+2} = g^6$
So inside: $12g^6$
Square it: $(12g^6)^2 = 144 \cdot g^{12} = 144g^{12}$

4) $(4hr^2)^2$
= $4^2 \cdot h^2 \cdot (r^2)^2 = 16 \cdot h^2 \cdot r^4 = 16h^2r^4$

5) $(4w^3 \cdot w^2 \cdot 3w)^2$
Coefficients: $4 \cdot 3 = 12$
Powers of $w$: $w^3 \cdot w^2 \cdot w^1 = w^{3+2+1} = w^6$
Inside: $12w^6$
Square: $144w^{12}$

6) $(4z^3 \cdot 3z \cdot z^2)^2$
Coefficients: $4 \cdot 3 = 12$
Powers: $z^3 \cdot z^1 \cdot z^2 = z^{6}$
Inside: $12z^6$
Square: $144z^{12}$

7) $(2d^4y)^2$
= $2^2 \cdot (d^4)^2 \cdot y^2 = 4 \cdot d^8 \cdot y^2 = 4d^8y^2$

8) $(4c^2 \cdot c^3)^2$
Inside: $4 \cdot c^{2+3} = 4c^5$
Square: $16c^{10}$

9) $(g \cdot 2g^2)^3$
Inside: $2 \cdot g^{1+2} = 2g^3$
Cube: $2^3 \cdot (g^3)^3 = 8 \cdot g^9 = 8g^9$

10) $(4n^2 \cdot n)^3$
Inside: $4 \cdot n^{2+1} = 4n^3$
Cube: $64 \cdot n^9 = 64n^9$

11) $(s \cdot 3s^3 \cdot s^3)^2$
Coefficients: $3$
Powers: $s^1 \cdot s^3 \cdot s^3 = s^{7}$
Inside: $3s^7$
Square: $9s^{14}$

12) $(3d^2 \cdot 4d)^3$
Coefficients: $3 \cdot 4 = 12$
Powers: $d^2 \cdot d^1 = d^3$
Inside: $12d^3$
Cube: $12^3 = 1728$, and $(d^3)^3 = d^9$ → $1728d^9$

13) $(2z^2 \cdot 3z^3 \cdot z)^2$
Coefficients: $2 \cdot 3 = 6$
Powers: $z^2 \cdot z^3 \cdot z^1 = z^{6}$
Inside: $6z^6$
Square: $36z^{12}$

14) $(3y^2 \cdot y \cdot 4)^3$
Coefficients: $3 \cdot 4 = 12$
Powers: $y^2 \cdot y^1 = y^3$
Inside: $12y^3$
Cube: $12^3 = 1728$, and $(y^3)^3 = y^9$ → $1728y^9$

Let me double-check a few tricky ones:

- #12: $3d^2 \cdot 4d = 12d^3$, cube → $12^3 = 1728$, $d^{3\cdot3}=d^9$ ✔
- #14: $3y^2 \cdot y = 3y^3$, times 4 = $12y^3$, cube = $1728y^9$ ✔
- #11: $s \cdot s^3 \cdot s^3 = s^7$, times 3 = $3s^7$, square = $9s^{14}$ ✔

All look correct.

Final Answer:
1) $216w^{18}$
2) $36h^8$
3) $144g^{12}$
4) $16h^2r^4$
5) $144w^{12}$
6) $144z^{12}$
7) $4d^8y^2$
8) $16c^{10}$
9) $8g^9$
10) $64n^9$
11) $9s^{14}$
12) $1728d^9$
13) $36z^{12}$
14) $1728y^9$