Question

Simplify the following: (7/8)^-3 x (9/5)^0 x 8^-2 x(1/7)^-1​

Answer 1

$\\ (\frac{7}{8} ) {}^{ - 3} \times (\frac{9}{5} ){}^{0} \times {8}^{ - 2} \times (\frac{1}{7} ) {}^{ - 1} \\ = > (\frac{8}{7} ){}^{3} \times 1 \times \frac{1}{64} \times 7 \\ = > \frac{8 \times 64 \times 7}{49 \times 7 \times 64} \\ = > \frac{8}{49}$

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