Multiply, and then simplify if possible. Assume that all variables represent positive real numbers. \( (\sqrt{7 x}-3 \sqrt{3 x})(\sqrt{7 x}-4 \sqrt{3 x}) \) \( (\sqrt{7 x}-3 \sqrt{3 x})(\sqrt{7 x}-4 \sqrt{3 x})=\square \) (Simplify your answer. Type an exact answer, using radicals as needed. Do not factor.)

To multiply \( (\sqrt{7 x}-3 \sqrt{3 x})(\sqrt{7 x}-4 \sqrt{3 x}) \), we can use the distributive property (or FOIL method for binomials): 1. First, we multiply the first terms: \[ \sqrt{7x} \cdot \sqrt{7x} = 7x \] 2. Next, we multiply the outer terms: \[ \sqrt{7x} \cdot (-4\sqrt{3x}) = -4\sqrt{21x^2} \] 3. Then, we multiply the inner terms: \[ (-3\sqrt{3x}) \cdot \sqrt{7x} = -3\sqrt{21x^2} \] 4. Finally, we multiply the last terms: \[ (-3\sqrt{3x})(-4\sqrt{3x}) = 12(3x) = 36x \] Now, we combine all these results: \[ 7x - 4\sqrt{21x^2} - 3\sqrt{21x^2} + 36x \] Combining like terms: \[ (7x + 36x) + (-4\sqrt{21x^2} - 3\sqrt{21x^2}) = 43x - 7\sqrt{21x^2} \] Since \(\sqrt{21x^2} = \sqrt{21} \cdot \sqrt{x^2} = \sqrt{21} \cdot x\), we can substitute that back in: \[ = 43x - 7x\sqrt{21} \] Putting it all together, we get: \[ 43x - 7x\sqrt{21} \] Thus, the simplified answer is: \[ \boxed{43x - 7x\sqrt{21}} \]