Question

Factorise using suitable identities
$a {}^{2} + \frac{1}{a {}^{2} } + 2$
​

Answer 1

Answer:

in picture

Step-by-step explanation:

pls refer to image hope it helps

Answer 2

$\large\underline{\sf{Solution-}}$

Given expression is

$\red{\rm :\longmapsto\: {a}^{2} + \dfrac{1}{ {a}^{2} } + 2}$

can be rewritten as

$\rm \:  =  \: {a}^{2} + \dfrac{1}{ {a}^{2} } + 2 \times 1$

$\rm \:  =  \: {a}^{2} + \dfrac{1}{ {a}^{2} } + 2 \times a \times \dfrac{1}{a}$

$\rm \:  =  \: {(a)}^{2} + {\bigg[\dfrac{1}{a} \bigg]}^{2} + 2 \times a \times \dfrac{1}{a}$

We know,

$\red{ \boxed{ \sf{ \: {x}^{2} + {y}^{2} + 2xy = {(x + y)}^{2}}}}$

Here,

$\red{\rm :\longmapsto\:x = a}$

and

$\red{\rm :\longmapsto\:y = \dfrac{1}{a} }$

So, using this identity, it can be rewritten as

$\rm \:  =  \: {\bigg[a + \dfrac{1}{a} \bigg]}^{2}$

$\rm \:  =  \: \bigg[a + \dfrac{1}{a} \bigg] \times \bigg[1 + \dfrac{1}{a} \bigg]$

Hence,

$\rm :\longmapsto\:\green{ \boxed{ \sf{ \: {a}^{2} + \frac{1}{ {a}^{2} } + 2 = \bigg[a + \dfrac{1}{a} \bigg] \times \bigg[1 + \dfrac{1}{a} \bigg]}}}$

Alternative Method

Given expression is

$\red{\rm :\longmapsto\: {a}^{2} + \dfrac{1}{ {a}^{2} } + 2}$

$\rm \:  =  \: {a}^{2} + \dfrac{1}{ {a}^{2} } + 2 \times1$

$\rm \:  =  \: {a}^{2} + \dfrac{1}{ {a}^{2} } + \dfrac{2a}{a}$

$\rm \:  =  \: {a}^{2} + \dfrac{1}{ {a}^{2} } + \dfrac{a}{a} + \dfrac{a}{a}$

$\rm \:  =  \: \bigg( {a}^{2} + \dfrac{a}{a} \bigg) + \bigg(\dfrac{1}{ {a}^{2} } + \dfrac{a}{a} \bigg)$

$\rm \:  =  \: a\bigg(a + \dfrac{1}{a} \bigg) + \dfrac{1}{a} \bigg(\dfrac{1}{a} + a\bigg)$

$\rm \:  =  \: a\bigg(a + \dfrac{1}{a} \bigg) + \dfrac{1}{a} \bigg(a + \dfrac{1}{a} \bigg)$

$\rm \:  =  \: \bigg(a + \dfrac{1}{a} \bigg) \times \bigg(a + \dfrac{1}{a} \bigg)$

Hence,

$\rm :\longmapsto\:\red{ \boxed{ \sf{ \: {a}^{2} + \frac{1}{ {a}^{2} } + 2 = \bigg[a + \dfrac{1}{a} \bigg] \times \bigg[1 + \dfrac{1}{a} \bigg]}}}$