Question

# Simplify the following term and fetch the result$\dfrac{{{{\left( {{2^5}} \right)}^2} \times {7^3}}}{{{8^3} \times 7}}$

Given:$\dfrac{{{{\left( {{2^5}} \right)}^2} \times {7^3}}}{{{8^3} \times 7}}$
We know that$\left[ {{{\left( {{x^a}} \right)}^b} = {x^{a \times b}}\& \left( {{2^3}} \right) = 8} \right]$
$\dfrac{{{{\left( {{2^5}} \right)}^2} \times {7^3}}}{{{8^3} \times 7}} \\ = \dfrac{{{{\left( {{2^5}} \right)}^2} \times {7^3}}}{{{{\left( {{2^3}} \right)}^3} \times 7}} \\ = \dfrac{{\left( {{2^{5 \times 2}}} \right) \times {7^3}}}{{\left( {{2^{3 \times 3}}} \right) \times 7}} \\ = \dfrac{{{2^{10}} \times {7^3}}}{{{2^9} \times 7}} \\ = 2 \times {7^2} = 2 \times 49 = 98 \\$