Factorise: $9{{x}^{2}}-16{{y}^{2}}$.

Hint: Use the identity \[''\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)''\] to factorise the given algebraic expression. For using this identity first convert the given algebraic expression in the form ${{\left( a \right)}^{2}}-{{\left( b \right)}^{2}}$.

Complete step-by-step answer:

Given algebraic expression: $9{{x}^{2}}-16{{y}^{2}}............\left( 1 \right)$

Identity to be used: \[{{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right).........\left( 2 \right)\]

Proof of this identity:

$LHS={{a}^{2}}-{{b}^{2}}$

Property: If we add something to an expression and then subtract the same from the expression, the expression will not change.

By using this property, let’s add and subtract “ab” from the expression of LHS. So, LHS will be,

$LHS=\left( {{a}^{2}}-{{b}^{2}} \right)+\left( ab-ab \right)$

On rearranging the terms, we will get,

$LHS=\left( {{a}^{2}}-ab \right)+\left( ab-{{b}^{2}} \right)$

Now, taking ‘a’ common from the first two terms and ‘b’ common from the last two terms, we will get,

$LHS=a\left( a-b \right)+b\left( a-b \right)$

Now, taking $\left( a-b \right)$ common from the two terms, we will get,

$LHS=\left( a-b \right)\left( a+b \right)$

From equation (2), we can see that, $RHS=\left( a-b \right)\left( a+b \right)$.

Hence, we proved this identity,

i.e. \[{{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)\]

Now, using this identity to factorise given algebraic expression $9{{x}^{2}}-16{{y}^{2}}$.

First we need to change this expression into the form ${{\left( a \right)}^{2}}-{{\left( b \right)}^{2}}$.

We know,

$\begin{align}

  & {{\left( 3x \right)}^{2}}=9{{x}^{2}}\ and \\

 & {{\left( 4x \right)}^{2}}=16{{x}^{2}} \\

\end{align}$

So, $9{{x}^{2}}-16{{y}^{2}}$ can be written as ${{\left( 3x \right)}^{2}}-{{\left( 4y \right)}^{2}}$.
 $i.e.\ 9{{x}^{2}}-16{{y}^{2}}={{\left( 3x \right)}^{2}}-{{\left( 4y \right)}^{2}}...........\left( 3 \right)$

Using the identity, ${{\left( 3x \right)}^{2}}-{{\left( 4y \right)}^{2}}=\left( 3x-4y \right)\left( 3x+4y \right)$

$\Rightarrow \left( 3x-4y \right)\left( 3x+4y \right)={{\left( 3x \right)}^{2}}-{{\left( 4y \right)}^{2}}..........\left( 4 \right)$

In equation (3) and (4), we can see that RHS of both the equations are same and equal to ${{\left( 3x \right)}^{2}}-{{\left( 4y \right)}^{2}}$. So, the LHS of the two equations will also be the same.

$\Rightarrow \left( 9{{x}^{2}}-16{{y}^{2}} \right)=\left( 3x-4y \right)\left( 3x+4y \right)$


Note: If you don’t remember the identity, you can solve it by using “middle term splitting’ method. If an expression in the form $''a{{x}^{2}}+bxy+c{{y}^{2}}''$, we split its middle term in such a way that two values got from splitting the middle term, when added together produces $bxy$ and when multiplied together produces $\left( a{{x}^{2}} \right)\times \left( c{{y}^{2}} \right)$.

Here, our expression is, $9{{x}^{2}}-16{{y}^{2}}$. Comparing it with $a{{x}^{2}}+bxy+c{{y}^{2}}=0$, we will get,

$a=9,b=0\ and\ c=-16$.

We need two terms, which when added together give ‘0’ and when multiplied together give $\left( 9{{x}^{2}} \right)\times \left( -16{{y}^{2}} \right)$.

The two terms will be, $-12xy\ and\ +12xy$.

As, $-12xy\ +12xy=0\ and$

$\left( -12xy \right)\ \left( 12xy \right)=-14{{x}^{2}}{{y}^{2}}=\left( 9{{x}^{2}} \right)\left( -16{{y}^{2}} \right)$

Now, $9{{x}^{2}}-16{{y}^{2}}$ can be written as $9{{x}^{2}}-12xy+12xy+16{{y}^{2}}$.

$i.e.\ 9{{x}^{2}}-16{{y}^{2}}=9{{x}^{2}}-12xy+12xy-16{{y}^{2}}$

Taking common: \[3x\left( 3x-4y \right)+4y\left( 3x-4y \right)\]

\[9{{x}^{2}}-16{{y}^{2}}=\left( 3x+4y \right)\left( 3x-4y \right)\]