### Instruction: Calculate $\sqrt{75x} \cdot \sqrt{2x} \cdot \sqrt{14x}$ . Express your answer in simplest radical form in terms of $x$.### Response: To simplify this product of square roots, I can use the property that $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$ for any non-negative numbers $a$ and $b$.
Applying this property, I can rewrite the product as $\sqrt{75x \cdot 2x \cdot 14x}$ .
I can simplify this further by multiplying the coefficients and the variables separately.
For the coefficients, I have $75 \cdot 2 \cdot 14 = 150 \cdot 14 = 2100$.
For the variables, I have $x \cdot x \cdot x = x^3$.
So the product is $\sqrt{2100x^3}$ .
I can simplify this even more by factoring out perfect squares from the radicand.
I notice that $2100 = 100 \cdot 21$ and $100 = 10^2$ is a perfect square.
I also notice that $x^3 = x^2 \cdot x$ and $x^2$ is a perfect square.
So I can rewrite the product as $\sqrt{100x^2} \cdot \sqrt{21x}$ .
Using the property that $\sqrt{a^2} = a$ for any non-negative number $a$, I can simplify the first factor as $10x$.
So the product is $10x \cdot \sqrt{21x}$ .
I cannot simplify this any further, so this is the final answer.