### Instruction: Simplify $\sqrt{28x} \cdot \sqrt{15x} \cdot \sqrt{21x}$. Express your answer in simplest radical form in terms of $x$.

Note: When entering a square root with more than one character, you must use parentheses or brackets.  For example, you should enter $\sqrt{14}$ as "sqrt(14)" or "sqrt{14}".### Response: I notice that all three factors have the same variable $x$ inside the square root, so I can simplify them by using the property that $\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$.
Applying this property to each factor, I get $\sqrt{28x} \cdot \sqrt{15x} \cdot \sqrt{21x} = \sqrt{28} \cdot \sqrt{x} \cdot \sqrt{15} \cdot \sqrt{x} \cdot \sqrt{21} \cdot \sqrt{x}$.
Now I can group the terms with $x$ together and use another property of square roots: $\sqrt{a} \cdot \sqrt{a} = \sqrt{a^2} = a$.
This gives me $\sqrt{28} \cdot \sqrt{x} \cdot \sqrt{15} \cdot \sqrt{x} \cdot \sqrt{21} \cdot \sqrt{x} = \sqrt{28} \cdot \sqrt{15} \cdot \sqrt{21} \cdot \sqrt{x^3}$.
I can simplify this further by looking for perfect square factors in each radicand (the number inside the square root).
For $\sqrt{28}$, I see that $28 = 4 \cdot 7$, and $4$ is a perfect square.
So I can write $\sqrt{28} = \sqrt{4} \cdot \sqrt{7} = 2 \cdot \sqrt{7}$.
For $\sqrt{15}$, I don't see any perfect square factors, so I leave it as it is.
For $\sqrt{21}$, I also don't see any perfect square factors, so I leave it as it is.
For $\sqrt{x^3}$, I notice that $x^3 = x^2 \cdot x$, and $x^2$ is a perfect square.
So I can write $\sqrt{x^3} = \sqrt{x^2} \cdot \sqrt{x} = x \cdot \sqrt{x}$.
Putting these together, I get $\sqrt{28} \cdot \sqrt{15} \cdot \sqrt{21} \cdot \sqrt{x^3} = 2 \cdot \sqrt{7} \cdot \sqrt{15} \cdot \sqrt{21} \cdot x \cdot \sqrt{x}$.
I can simplify this even more by multiplying the constants and the variables together, and combining the square roots into one.
This gives me $2 \cdot \sqrt{7} \cdot \sqrt{15} \cdot \sqrt{21} \cdot x \cdot \sqrt{x} = 2x \cdot \sqrt{7 \cdot 15 \cdot 21 \cdot x}$.
Finally, I can check if there are any more perfect square factors in the radicand.
I see that $7 \cdot 15 \cdot 21 \cdot x = 7 \cdot 3 \cdot 5 \cdot 7 \cdot 3 \cdot x = 3^2 \cdot 7^2 \cdot 5 \cdot x$.
So I can write $\sqrt{7 \cdot 15 \cdot 21 \cdot x} = \sqrt{3^2 \cdot 7^2 \cdot 5 \cdot x} = 3 \cdot 7 \cdot \sqrt{5 \cdot x} = 21 \cdot \sqrt{5 \cdot x}$.
Therefore, the final answer is $2x \cdot \sqrt{7 \cdot 15 \cdot 21 \cdot x} = 2x \cdot 21 \cdot \sqrt{5 \cdot x} = 42x \cdot \sqrt{5 \cdot x}$.