To simplify the expression \( \sqrt{18 x^4 y^2 z} \times \sqrt{9 x^3 y^3 z} \), we can use the property of square roots that states \( \sqrt{a} \times \sqrt{b} = \sqrt{a \times b} \).
So, we have:
\[ \sqrt{18 x^4 y^2 z} \times \sqrt{9 x^3 y^3 z} = \sqrt{(18 x^4 y^2 z)(9 x^3 y^3 z)} \]
Now, let's first multiply the constants and then the variables inside the square root:
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Multiply the constants: \[ 18 \times 9 = 162 \]
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Combine the variables:
- For \(x\): \(x^4 \times x^3 = x^{4+3} = x^7\)
- For \(y\): \(y^2 \times y^3 = y^{2+3} = y^5\)
- For \(z\): \(z \times z = z^{1+1} = z^2\)
Putting it all together gives us:
\[ \sqrt{162 x^7 y^5 z^2} \]
Next, let's simplify the square root. We can break down \(162\) and look at the other variables:
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Prime factorization of \(162\): \[ 162 = 81 \times 2 = 9^2 \times 2 = 3^4 \times 2 \]
Thus \(162 = 9 \times 18 = 9 \times (9 \times 2) = 81 \times 2\).
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Separate the perfect squares: \[ \sqrt{162} = \sqrt{81 \cdot 2} = \sqrt{81} \times \sqrt{2} = 9\sqrt{2} \]
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Extract squares from the variables:
- \(x^7 = x^{6} \cdot x = (x^3)^2 \cdot x \) gives \( x^3 \) outside the square root.
- \(y^5 = y^{4} \cdot y = (y^2)^2 \cdot y\) gives \( y^2 \) outside the square root.
- \(z^2 = (z)^2\) gives \( z \) outside the square root.
Now, putting it altogether gives us:
\[ \sqrt{162 x^7 y^5 z^2} = 9 \sqrt{2} \cdot x^{3} \cdot y^{2} \cdot z = 9 x^3 y^2 z \sqrt{2} \]
Thus, the final simplified result is:
\[ \boxed{9 x^3 y^2 z \sqrt{2}} \]