Question

Simplify the given algebraic expression. $14xy\left( {81{x^2} - 121{y^2}} \right) \div 7xy(9x + 11y)$

Answer 1

Hint: We simplify the given expression by taking common terms in the numerator and the denominator, then we will use property like \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\] for further simplification.
Now, after the simplification, we will get the required expression as it will be the simplest form of the given expression.

Complete step-by-step answer:
Given data: $14xy\left( {81{x^2} - 121{y^2}} \right) \div 7xy(9x + 11y)$
We know that if the division of any number is done by itself than the result is always 1.
Therefore, solving for the expressions i.e. $14xy\left( {81{x^2} - 121{y^2}} \right) \div 7xy(9x + 11y)$
 $ \Rightarrow \dfrac{{14xy\left( {81{x^2} - 121{y^2}} \right)}}{{7xy(9x + 11y)}}$
Taking 7xy common from both the denominator and the numerator
 $ \Rightarrow \dfrac{{7xy}}{{7xy}}\left( {\dfrac{{2\left( {81{x^2} - 121{y^2}} \right)}}{{(9x + 11y)}}} \right)$
On simplifying we get,
 $ \Rightarrow \dfrac{{2\left( {81{x^2} - 121{y^2}} \right)}}{{(9x + 11y)}}$
Using \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\] , we get,
\[ \Rightarrow \] $\dfrac{{2\left( {9x - 11y} \right)\left( {9x + 11y} \right)}}{{\left( {9x + 11y} \right)}}$
On cancelling common terms we get,
 $ \Rightarrow 2\left( {9x - 11y} \right)$.

Note: In the above solution we have used a property of algebra that is \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\],
We can also prove this property as solving the right-hand side expression of the equation
Taking the right-hand side expression of the equation, we get
  \[ = \left( {a + b} \right)\left( {a - b} \right)\]
On multiplying the bracket terms and simplifying
 \[ = {a^2} - ab + ab - {b^2}\]
Since the sum of numbers have equal magnitude but the opposite sign is zero
Hence, we get, ${a^2} - {b^2}$ , which is equal to the expression in the left-hand side of the equation