Solve for \( x \) \( \left(\frac{1}{3}\right)^{3 x+3}=729 \)

To solve the equation \( \left(\frac{1}{3}\right)^{3 x+3}=729 \), we start by expressing 729 as a power of 3. We know that \( 729 = 3^6 \). This allows us to rewrite the equation as: \[ \left(\frac{1}{3}\right)^{3 x + 3} = 3^6 \] Next, we can write \( \frac{1}{3} \) as \( 3^{-1} \): \[ (3^{-1})^{3 x + 3} = 3^6 \] Using the power of a power property, we can simplify the left side: \[ 3^{-(3 x + 3)} = 3^6 \] Now, we can equate the exponents since the bases are the same: \[ -(3 x + 3) = 6 \] This simplifies to: \[ -3 x - 3 = 6 \] Adding 3 to both sides gives: \[ -3 x = 9 \] Dividing both sides by -3 results in: \[ x = -3 \] Thus, the solution to the equation is: \[ \boxed{-3} \]