$$ \begin{array}{ll}\frac{7^{6}}{7^{4}} & ext { ii) } \frac{2^{4} \cdot 5^{3}}{10^{2}}\end{array} \quad $$ iii) $$ \left\{\frac{(a+b)^{2} \cdot(c+d)^{3}}{(a+b) \cdot(c+d)^{2}} ight\}^{3} $$

Question

$$ \begin{array}{ll}\frac{7^{6}}{7^{4}} & 	ext { ii) } \frac{2^{4} \cdot 5^{3}}{10^{2}}\end{array} \quad $$ iii) $$ \left\{\frac{(a+b)^{2} \cdot(c+d)^{3}}{(a+b) \cdot(c+d)^{2}}ight\}^{3} $$

UpStudy AI · Accepted Answer

Let's simplify each expression step by step.

### i) $$ \frac{7^{6}}{7^{4}} $$

Using the property of exponents that states $$ \frac{a^m}{a^n} = a^{m-n} $$:

$$
\frac{7^{6}}{7^{4}} = 7^{6-4} = 7^{2}
$$

So, the simplified result is:

$$
7^{2} = 49
$$

### ii) $$ \frac{2^{4} \cdot 5^{3}}{10^{2}} $$

First, we can express $$ 10^{2} $$ in terms of its prime factors:

$$
10^{2} = (2 \cdot 5)^{2} = 2^{2} \cdot 5^{2}
$$

Now we can rewrite the expression:

$$
\frac{2^{4} \cdot 5^{3}}{10^{2}} = \frac{2^{4} \cdot 5^{3}}{2^{2} \cdot 5^{2}}
$$

Using the property of exponents again:

$$
= \frac{2^{4}}{2^{2}} \cdot \frac{5^{3}}{5^{2}} = 2^{4-2} \cdot 5^{3-2} = 2^{2} \cdot 5^{1}
$$

Calculating the powers:

$$
2^{2} = 4 \quad \text{and} \quad 5^{1} = 5
$$

Thus, we have:

$$
2^{2} \cdot 5^{1} = 4 \cdot 5 = 20
$$

### iii) $$ \left\{\frac{(a+b)^{2} \cdot(c+d)^{3}}{(a+b) \cdot(c+d)^{2}}\right\}^{3} $$

First, simplify the fraction inside the braces:

$$
\frac{(a+b)^{2} \cdot(c+d)^{3}}{(a+b) \cdot(c+d)^{2}} = \frac{(a+b)^{2}}{(a+b)} \cdot \frac{(c+d)^{3}}{(c+d)^{2}} = (a+b)^{2-1} \cdot (c+d)^{3-2} = (a+b)^{1} \cdot (c+d)^{1}
$$

This simplifies to:

$$
(a+b)(c+d)
$$

Now, we raise this result to the power of 3:

$$
\left\{(a+b)(c+d)\right\}^{3} = (a+b)^{3} \cdot (c+d)^{3}
$$

Thus, the final result is:

$$
(a+b)^{3} \cdot (c+d)^{3}
$$

### Summary of Results

1. $$ \frac{7^{6}}{7^{4}} = 49 $$
2. $$ \frac{2^{4} \cdot 5^{3}}{10^{2}} = 20 $$
3. $$ \left\{\frac{(a+b)^{2} \cdot(c+d)^{3}}{(a+b) \cdot(c+d)^{2}}\right\}^{3} = (a+b)^{3} \cdot (c+d)^{3} $$
1. $$ \frac{7^{6}}{7^{4}} = 49 $$ 2. $$ \frac{2^{4} \cdot 5^{3}}{10^{2}} = 20 $$ 3. $$ \left\{\frac{(a+b)^{2} \cdot(c+d)^{3}}{(a+b) \cdot(c+d)^{2}}\right\}^{3} = (a+b)^{3} \cdot (c+d)^{3} $$

iii)

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