If $x$, $y$, $z$ are nonnegative integers satisfying the equation below, then compute $x+y+z$.
\[\left(\frac{16}{3}\right)^x\times \left(\frac{27}{25}\right)^y\times \left(\frac{5}{4}\right)^z=256.\]

[i]Proposed by Jeffrey Shi[/i]
1. Start with the given equation:
   \[
   \left(\frac{16}{3}\right)^x \times \left(\frac{27}{25}\right)^y \times \left(\frac{5}{4}\right)^z = 256
   \]

2. Rewrite each fraction in terms of prime factors:
   \[
   \frac{16}{3} = \frac{2^4}{3}, \quad \frac{27}{25} = \frac{3^3}{5^2}, \quad \frac{5}{4} = \frac{5}{2^2}
   \]

3. Substitute these into the equation:
   \[
   \left(\frac{2^4}{3}\right)^x \times \left(\frac{3^3}{5^2}\right)^y \times \left(\frac{5}{2^2}\right)^z = 256
   \]

4. Simplify the equation:
   \[
   \frac{2^{4x}}{3^x} \times \frac{3^{3y}}{5^{2y}} \times \frac{5^z}{2^{2z}} = 256
   \]

5. Combine the terms:
   \[
   \frac{2^{4x} \times 3^{3y} \times 5^z}{3^x \times 5^{2y} \times 2^{2z}} = 256
   \]

6. Simplify the exponents:
   \[
   2^{4x-2z} \times 3^{3y-x} \times 5^{z-2y} = 2^8 \times 3^0 \times 5^0
   \]

7. Equate the exponents of the corresponding prime factors:
   \[
   4x - 2z = 8 \quad (1)
   \]
   \[
   3y - x = 0 \quad (2)
   \]
   \[
   z - 2y = 0 \quad (3)
   \]

8. Solve the system of equations:
   - From equation (2):
     \[
     x = 3y
     \]

   - From equation (3):
     \[
     z = 2y
     \]

   - Substitute \( x = 3y \) and \( z = 2y \) into equation (1):
     \[
     4(3y) - 2(2y) = 8
     \]
     \[
     12y - 4y = 8
     \]
     \[
     8y = 8
     \]
     \[
     y = 1
     \]

   - Using \( y = 1 \):
     \[
     x = 3y = 3 \times 1 = 3
     \]
     \[
     z = 2y = 2 \times 1 = 2
     \]

9. Therefore, the values of \( x, y, z \) are:
   \[
   (x, y, z) = (3, 1, 2)
   \]

10. Compute \( x + y + z \):
    \[
    x + y + z = 3 + 1 + 2 = 6
    \]

The final answer is \(\boxed{6}\)