Find the product. \( (2 r-3 t)^{3} \)

To find the product \( (2r - 3t)^3 \), we can use the binomial theorem, which states that \( (a - b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} (-b)^k \). Here, \( a = 2r \), \( b = 3t \), and \( n = 3 \). Applying the formula, we have: \[ (2r - 3t)^3 = \sum_{k=0}^{3} \binom{3}{k} (2r)^{3-k} (-3t)^k. \] Calculating each term: - For \( k = 0 \): \(\binom{3}{0} (2r)^3 (-3t)^0 = 1 \cdot 8r^3 \cdot 1 = 8r^3\) - For \( k = 1 \): \(\binom{3}{1} (2r)^2 (-3t)^1 = 3 \cdot 4r^2 \cdot (-3t) = -36r^2t\) - For \( k = 2 \): \(\binom{3}{2} (2r)^1 (-3t)^2 = 3 \cdot 2r \cdot 9t^2 = 54rt^2\) - For \( k = 3 \): \(\binom{3}{3} (2r)^0 (-3t)^3 = 1 \cdot 1 \cdot (-27t^3) = -27t^3\) Now, combine all the terms: \[ (2r - 3t)^3 = 8r^3 - 36r^2t + 54rt^2 - 27t^3. \] So, the product is: \[ \boxed{8r^3 - 36r^2t + 54rt^2 - 27t^3}. \]