2. Show that a) $$ \sqrt[n]{a} \cdot \sqrt[n]{b}=\sqrt[n]{a b} $$

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To prove that

$$
\sqrt[n]{a} \cdot \sqrt[n]{b} = \sqrt[n]{a b},
$$

we can use the properties of exponents and radicals. Here's a step-by-step demonstration:

### Step 1: Express Radicals as Exponents

Recall that the $$n$$-th root of a number can be expressed as that number raised to the power of $$1/n$$. Therefore, we can rewrite the radicals as exponents:

$$
\sqrt[n]{a} = a^{1/n} \quad \text{and} \quad \sqrt[n]{b} = b^{1/n}.
$$

### Step 2: Multiply the Expressions

Now, multiply the two expressions:

$$
\sqrt[n]{a} \cdot \sqrt[n]{b} = a^{1/n} \cdot b^{1/n}.
$$

### Step 3: Use the Property of Exponents

When multiplying two expressions with the same exponent, you can multiply the bases and keep the exponent:

$$
a^{1/n} \cdot b^{1/n} = (a \cdot b)^{1/n}.
$$

### Step 4: Convert Back to Radical Notation

Finally, convert the expression back to radical form:

$$
(a \cdot b)^{1/n} = \sqrt[n]{a b}.
$$

### Conclusion

Putting it all together, we have shown that:

$$
\sqrt[n]{a} \cdot \sqrt[n]{b} = \sqrt[n]{a b}.
$$

This property is a fundamental rule of radicals and exponents, allowing us to combine roots of the same degree through multiplication.
$$ \sqrt[n]{a} \cdot \sqrt[n]{b} = \sqrt[n]{a b} $$

Show that
a)

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