### 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: To simplify this expression, I want to use the property that $\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$ for any positive numbers $a$ and $b$.
This allows me to combine the square roots under one radical sign.
Applying this property, I get $\sqrt{28x} \cdot \sqrt{15x} \cdot \sqrt{21x} = \sqrt{(28x)(15x)(21x)} = \sqrt{8820x^3}$.
Now, I want to simplify the radicand, which is the expression inside the square root.
To do this, I want to factor out any perfect squares that are factors of 8820.
I can use a prime factorization tree to find the prime factors of 8820.
Starting from 8820, I can divide by 2 to get 4410, then by 2 again to get 2205, then by 3 to get 735, then by 3 again to get 245, then by 5 to get 49, then by 7 to get 7.
This means that 8820 = 2^2 \cdot 3^2 \cdot 5 \cdot 7^2.
I can also see that $x^3 = x \cdot x \cdot x$ is a perfect square times $x$.
Therefore, I can rewrite the radicand as $\sqrt{8820x^3} = \sqrt{(2^2)(3^2)(5)(7^2)(x^2)(x)}$.
Now, I can use the property that $\sqrt{a^2} = a$ for any positive number $a$.
This allows me to take out any perfect squares from under the square root and multiply them outside.
Doing this, I get $\sqrt{(2^2)(3^2)(5)(7^2)(x^2)(x)} = (2)(3)(7)(x) \cdot \sqrt{5x} = 42x \cdot \sqrt{5x}$.
This is the simplest radical form of the expression.