Answer: The correct option is (C) ![8x^4yz^4\sqrt{2yz}.](https://tex.z-dn.net/?f=8x%5E4yz%5E4%5Csqrt%7B2yz%7D.)
Step-by-step explanation: We are given to select the correct expression that is equivalent to the expression below:
![E=\sqrt{128x^8y^3z^9}.](https://tex.z-dn.net/?f=E%3D%5Csqrt%7B128x%5E8y%5E3z%5E9%7D.)
We will be using the following properties of exponents:
![(i)~\sqrt a=a^\frac{1}{2},\\\\(ii)~(ab)^x=a^xb^x,\\\\(iii)~(a^x)^y=a^{x\times y}.](https://tex.z-dn.net/?f=%28i%29~%5Csqrt%20a%3Da%5E%5Cfrac%7B1%7D%7B2%7D%2C%5C%5C%5C%5C%28ii%29~%28ab%29%5Ex%3Da%5Exb%5Ex%2C%5C%5C%5C%5C%28iii%29~%28a%5Ex%29%5Ey%3Da%5E%7Bx%5Ctimes%20y%7D.)
Since y > 0 and z > 0, so we will be considering the positive square root of y and z.
We have
![E\\\\=\sqrt{128x^8y^3z^9}\\\\=(128x^8y^3z^9)^\frac{1}{2}\\\\=(128)^\frac{1}{2}(x^8)^\frac{1}{2}(y^3)^\frac{1}{2}(z^9)^\frac{1}{2}\\\\=(2\times 8^2)^\frac{1}{2}x^{8\times \frac{1}{2}}y^{3\times\frac{1}{2}}z^{9\times\frac{1}{2}}\\\\=8x^4yz^4(2yz)^\frac{1}{2}\\\\=8x^4yz^4\sqrt{2yz}.](https://tex.z-dn.net/?f=E%5C%5C%5C%5C%3D%5Csqrt%7B128x%5E8y%5E3z%5E9%7D%5C%5C%5C%5C%3D%28128x%5E8y%5E3z%5E9%29%5E%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%3D%28128%29%5E%5Cfrac%7B1%7D%7B2%7D%28x%5E8%29%5E%5Cfrac%7B1%7D%7B2%7D%28y%5E3%29%5E%5Cfrac%7B1%7D%7B2%7D%28z%5E9%29%5E%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%3D%282%5Ctimes%208%5E2%29%5E%5Cfrac%7B1%7D%7B2%7Dx%5E%7B8%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%7Dy%5E%7B3%5Ctimes%5Cfrac%7B1%7D%7B2%7D%7Dz%5E%7B9%5Ctimes%5Cfrac%7B1%7D%7B2%7D%7D%5C%5C%5C%5C%3D8x%5E4yz%5E4%282yz%29%5E%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%3D8x%5E4yz%5E4%5Csqrt%7B2yz%7D.)
Thus, the equivalent expression is ![8x^4yz^4\sqrt{2yz}.](https://tex.z-dn.net/?f=8x%5E4yz%5E4%5Csqrt%7B2yz%7D.)
(C) is the correct option.