# Simplify the following term and fetch the result

$\dfrac{{{{\left( {{2^5}} \right)}^2} \times {7^3}}}{{{8^3} \times 7}}$

Answer

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Hint: Solve by using algebraic formulas, not by vigorous calculation.

Given:$\dfrac{{{{\left( {{2^5}} \right)}^2} \times {7^3}}}{{{8^3} \times 7}}$

We know that$\left[ {{{\left( {{x^a}} \right)}^b} = {x^{a \times b}}\& \left( {{2^3}} \right) = 8} \right]$

So continuing further

$

\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 \\

$

Hence, 98 is the solution.

Note:In such a type of problem which contains lots of numerical values, never try to simplify the terms by doing large calculations as that will be very tedious. Try to solve by cancelling some terms and then proceeding further.

Given:$\dfrac{{{{\left( {{2^5}} \right)}^2} \times {7^3}}}{{{8^3} \times 7}}$

We know that$\left[ {{{\left( {{x^a}} \right)}^b} = {x^{a \times b}}\& \left( {{2^3}} \right) = 8} \right]$

So continuing further

$

\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 \\

$

Hence, 98 is the solution.

Note:In such a type of problem which contains lots of numerical values, never try to simplify the terms by doing large calculations as that will be very tedious. Try to solve by cancelling some terms and then proceeding further.

Last updated date: 30th Sep 2023

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