7. \( \frac{21^{3}}{\left(27^{-2} \times 49^{6}\right)^{\frac{1}{3}}}= \)

To solve the expression \( \frac{21^{3}}{\left(27^{-2} \times 49^{6}\right)^{\frac{1}{3}}} \), we start by simplifying the denominator. First, rewrite \( 27 \) and \( 49 \) in terms of their prime factors: - \( 27 = 3^3 \), so \( 27^{-2} = (3^3)^{-2} = 3^{-6} \). - \( 49 = 7^2 \), so \( 49^6 = (7^2)^6 = 7^{12} \). Now, substituting back, we get: \[ 27^{-2} \times 49^{6} = 3^{-6} \times 7^{12} \] Next, we take the cube root of the entire expression: \[ \left(27^{-2} \times 49^{6}\right)^{\frac{1}{3}} = \left(3^{-6} \times 7^{12}\right)^{\frac{1}{3}} = 3^{-2} \times 7^{4} \] Now our expression becomes: \[ \frac{21^{3}}{3^{-2} \times 7^{4}} \] Rewriting \( 21 \) in terms of its prime factors: \[ 21 = 3^1 \times 7^1 \implies 21^3 = (3^1 \times 7^1)^3 = 3^{3} \times 7^{3} \] Now substitute back into the expression: \[ \frac{3^{3} \times 7^{3}}{3^{-2} \times 7^{4}} = 3^{3 - (-2)} \times 7^{3 - 4} = 3^{3 + 2} \times 7^{-1} = 3^{5} \times 7^{-1} \] This can be rewritten as: \[ \frac{3^{5}}{7} \] Now calculate \( 3^{5} \): \[ 3^{5} = 243 \] So the final answer is: \[ \frac{243}{7} \] Therefore, \( \frac{21^{3}}{\left(27^{-2} \times 49^{6}\right)^{\frac{1}{3}}} = \frac{243}{7} \).