Suppose that $x$ is real number such that $\frac{27\times 9^x}{4^x}=\frac{3^x}{8^x}$. Find the value of $2^{-(1+\log_23)x}$
1. Start with the given equation:
   \[
   \frac{27 \times 9^x}{4^x} = \frac{3^x}{8^x}
   \]

2. Express all terms with the same base:
   \[
   27 = 3^3, \quad 9^x = (3^2)^x = 3^{2x}, \quad 4^x = (2^2)^x = 2^{2x}, \quad 8^x = (2^3)^x = 2^{3x}
   \]

3. Substitute these expressions into the equation:
   \[
   \frac{3^3 \times 3^{2x}}{2^{2x}} = \frac{3^x}{2^{3x}}
   \]

4. Simplify the left-hand side:
   \[
   \frac{3^{3+2x}}{2^{2x}} = \frac{3^x}{2^{3x}}
   \]

5. Combine the exponents on the left-hand side:
   \[
   \frac{3^{3+2x}}{2^{2x}} = 3^x \cdot 2^{-3x}
   \]

6. Equate the exponents of the same bases on both sides:
   \[
   3^{3+2x} \cdot 2^{-2x} = 3^x \cdot 2^{-3x}
   \]

7. Separate the terms involving \(3\) and \(2\):
   \[
   3^{3+2x} = 3^x \quad \text{and} \quad 2^{-2x} = 2^{-3x}
   \]

8. Solve the exponents of \(3\):
   \[
   3 + 2x = x \implies 2x - x = -3 \implies x = -3
   \]

9. Verify the solution by substituting \(x = -3\) back into the original equation:
   \[
   \frac{27 \times 9^{-3}}{4^{-3}} = \frac{3^{-3}}{8^{-3}}
   \]
   Simplify:
   \[
   \frac{27 \times 3^{-6}}{2^{-6}} = \frac{3^{-3}}{2^{-9}}
   \]
   \[
   \frac{27 \times 3^{-6}}{2^{-6}} = 3^{-3} \cdot 2^9
   \]
   \[
   \frac{27}{3^6} \cdot 2^6 = 3^{-3} \cdot 2^9
   \]
   \[
   \frac{27}{729} \cdot 64 = \frac{1}{27} \cdot 512
   \]
   \[
   \frac{64}{27} = \frac{512}{27}
   \]
   Both sides are equal, confirming \(x = -3\) is correct.

10. Now, find the value of \(2^{-(1+\log_2 3)x}\):
    \[
    2^{-(1+\log_2 3)x} = 2^{-(1+\log_2 3)(-3)}
    \]
    \[
    = 2^{3(1+\log_2 3)}
    \]
    \[
    = 2^{3 + 3\log_2 3}
    \]
    \[
    = 2^3 \cdot 2^{3\log_2 3}
    \]
    \[
    = 8 \cdot 3^3
    \]
    \[
    = 8 \cdot 27
    \]
    \[
    = 216
    \]

The final answer is \(\boxed{216}\)