Question

\[ \frac{3^{-3} \times 6^2 \times \sqrt{98}} {5^2 \times \sqrt[3]{\frac{1}{25}} \times 15^{-4/3} \times 3^{1/3}} \]

Solution

\[ 6^2 = 2^2 \times 3^2,\quad \sqrt{98} = 7\sqrt{2},\quad 15 = 3 \times 5 \] \[ \sqrt[3]{\frac{1}{25}} = \frac{1}{5^{2/3}} \] \[ = \frac{3^{-3} \times 2^2 \times 3^2 \times 7\sqrt{2}} {5^2 \times \frac{1}{5^{2/3}} \times (3 \times 5)^{-4/3} \times 3^{1/3}} \] \[ = \frac{3^{-1} \times 2^2 \times 7\sqrt{2}} {5^2 \times 5^{-2/3} \times 5^{-4/3} \times 3^{-4/3} \times 3^{1/3}} \] \[ = \frac{3^{-1} \times 2^2 \times 7\sqrt{2}} {5^{2 – 2/3 – 4/3} \times 3^{-4/3 + 1/3}} \] \[ = \frac{3^{-1} \times 2^2 \times 7\sqrt{2}} {5^0 \times 3^{-1}} \] \[ = 3^{-1} \times 3^1 \times 2^2 \times 7\sqrt{2} \] \[ = 2^2 \times 7\sqrt{2} \] \[ = 28\sqrt{2} \]

Answer

\[ \boxed{28\sqrt{2}} \]

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