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Class 9
Maths
CBSE



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Answer 1

Question :-

Factorize $\longrightarrow \sf x^{4} + \dfrac{1}{x^{4}} + 1$

Answer :-

• In this question, no number has a common factor except 1

Notice that all the numbers are perfect squares.

$\implies \sf (x^{2})^{2} + \dfrac{1}{(x^{2})^{2}} + 1$

Expressing 1 as 2 - 1,

$\implies \sf (x^{2}) + \dfrac{1}{(x^{2})^{2}} + 2 - 1$

• Applying identity (a+b)² = a² + b² + 2ab

$\implies \sf \bigg [(x^{2})^{2} + \dfrac{1}{(x^{2})^{2}} + 2(x^{2} )\bigg(\dfrac{1}{x^{2} } \bigg)\bigg] - 1$

$\implies \sf \bigg( x^{2} + \dfrac{1}{x^{2} } \bigg)^{2} - 1$

• Applying identity a² - b² = (a+b)(a-b)

$\implies \sf \bigg(x^{2} + \dfrac{1}{x^{2} } \bigg)^{2} - (1)^{2}$

$\implies \sf \bigg[\bigg(x^{2} + \dfrac{1}{x^{2} }\bigg) + 1\bigg]\bigg[\bigg(x^{2} + \dfrac{1}{x^{2} } \bigg)-1\bigg]$

Here expressing $\sf \bigg[\bigg(x^{2} + \dfrac{1}{x^{2} } \bigg) - 1\bigg]$  the same way,

$\implies \sf \bigg[\bigg(x^{2} + \dfrac{1}{x^{2} } \bigg) + 1\bigg]\bigg[\bigg(x^{2} + \dfrac{1}{x^{2} } \bigg)+ 2 - 1\bigg]$

$\implies \sf \bigg[\bigg(x^{2} + \dfrac{1}{x^{2} } \bigg) + 1\bigg]\bigg[\bigg[x^{2} + \dfrac{1}{x^{2} y} + 2(x)\bigg(\dfrac{1}{x^{2} } \bigg)\bigg]- 1\bigg]$

$\implies \sf \bigg[\bigg(x^{2} + \dfrac{1}{x^{2} } \bigg) + 1\bigg]\bigg[\bigg(x + \dfrac{1}{x} \bigg)^{2} - 1\bigg]$

$\implies \sf \bigg[\bigg(x^{2} + \dfrac{1}{x^{2} } \bigg) + 1\bigg]\bigg[\bigg(x+ \dfrac{1}{x} \bigg) - 1\bigg]\bigg[\bigg(x + \dfrac{1}{x} \bigg)+1\bigg]$

Identities :-

• (a+b)² = a² + 2ab + b²
• (a-b)² = a² - 2ab + b²
• (a+b+c)² = a² + b² + c² + 2ab + 2bc + 2ca
• (x+a)(x+b) = x² + x(a+b) + ab
• a²-b² = (a+b)(a-b)
• (a+b)³ = a³ + 3a²b + 3ab² + b³
• (a-b)³ = a³ - 3a²b + 3ab² - b³
• a³+b³ = (a+b)(a² - ab + b²)
• a³-b³ = (a-b)(a² + ab + b²)
• a³+b³+c³ - 3abc = (a+b+c)(a² + b² + c² - ab - bc - ca)

Answer 2

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

$\rm :\longmapsto\: {x}^{4} + \dfrac{1}{ {x}^{4} } + 1$

$\sf \: = \: \: \dfrac{ {x}^{8} + 1 + {x}^{4} }{ {x}^{4} }$

$\sf \: = \: \: \dfrac{1}{ {x}^{4} } \bigg( {x}^{8} + 2{x}^{4} + 1 - {x }^{4} \bigg)$

$\: \: \: \: \: \: \: \: \: \: \: \because \boxed{ \sf \: on \: adding \: and \: subtracting \: {x}^{4} }$

$\sf \: = \: \: \dfrac{1}{ {x}^{4} } \bigg( {( {x}^{4} + 1) }^{2} - { ({x}^{2}) }^{2} \bigg)$

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

$\sf \: = \: \: \dfrac{1}{ {x}^{4} } \bigg( {x}^{4} + 1 - {x}^{2} \bigg) \bigg( {x}^{4} + 1 + {x}^{2} \bigg)$

$\: \: \: \: \: \: \: \: \: \boxed{ \sf \: \because \: {x}^{2} - {y}^{2} = (x + y)(x - y)}$

$\sf \: = \: \: \dfrac{1}{ {x}^{4} } \bigg( {x}^{4} + 1 - {x}^{2} \bigg) \bigg( {x}^{4} + 1 + {2x}^{2} - {x}^{2} \bigg)$

$\: \: \: \: \: \: \: \: \: \: \: \because \boxed{ \sf \: on \: adding \: and \: subtracting \: {x}^{2} }$

$\sf \: = \: \: \dfrac{1}{ {x}^{4} } \bigg( {x}^{4} + 1 - {x}^{2} \bigg)\bigg( {( {x}^{2} + 1) }^{2} - {(x)}^{2} \bigg)$

$\sf \: = \: \: \dfrac{1}{ {x}^{4} } \bigg( {x}^{4} + 1 - {x}^{2} \bigg) \bigg(( {x}^{2} + 1 + x)( {x}^{2} + 1 - x) \bigg)$

$\sf \: = \: \: \bigg( \dfrac{ {x}^{4} + 1 - {x}^{2} }{ {x}^{2} } \bigg) \bigg( \dfrac{ {x}^{2} + x + 1 }{x} \bigg) \bigg( \dfrac{ {x}^{2} - x + 1}{x} \bigg)$

$\sf \: = \:\bigg( {x}^{2} + \dfrac{1}{ {x}^{2} } - 1 \bigg) \bigg( x + \dfrac{1}{x} + 1 \bigg) \bigg(x + \dfrac{1}{x} - 1 \bigg)$

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Additional Information :-

More Identities to know:

• (a + b)² = a² + 2ab + b²

• (a - b)² = a² - 2ab + b²

• a² - b² = (a + b)(a - b)

• (a + b)² = (a - b)² + 4ab

• (a - b)² = (a + b)² - 4ab

• (a + b)² + (a - b)² = 2(a² + b²)

• (a + b)³ = a³ + b³ + 3ab(a + b)

• (a - b)³ = a³ - b³ - 3ab(a - b)