Question

factorise :- x² + 1/x² - 2 - 3x + 3/x
In Details ​

Answer 1

$\large\underline{\bold{Given \:Question - }}$

$\sf \: Factorize : \: {x}^{2} + \dfrac{1}{ {x}^{2} } - 2 - 3x + \dfrac{3}{x}$

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

$\rm :\longmapsto\:\: {x}^{2} + \dfrac{1}{ {x}^{2} } - 2 - 3x + \dfrac{3}{x}$

$\: \: \: \: = \: \rm \: \: \: \bigg({x}^{2} + \dfrac{1}{ {x}^{2} } \bigg) - 3 \bigg(x - \dfrac{1}{x} \bigg) - 2$

$\bigg \{\sf \: Let \: x - \dfrac{1}{x} = y \\ \sf \: squaring \: both \: sides \\ \sf \: {\bigg(x - \dfrac{1}{x} \bigg) }^{2} = {y}^{2} \\ \sf \: {x}^{2} + \dfrac{1}{ {x}^{2} } - 2 = {y}^{2} \\ \sf \: {x}^{2} + \dfrac{1}{ {x}^{2}} = {y}^{2} + 2 \bigg \}$

$\: \: \: \: = \: \rm \: \: ({y}^{2} + 2) - 3y - 2$

$\: \: \: \: = \: \rm \: \: {y}^{2} + 2 - 3y - 2$

$\: \: \: \: = \: \rm \: \: {y}^{2} - 3y$

$\: \: \: \: = \: \rm \: \: y(y - 3)$

$\: \: \: \: = \: \rm \: \: \bigg(x - \dfrac{1}{x}\bigg) \bigg(x - \dfrac{1}{x} - 3 \bigg)$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \bf \because \: y \: = \: x - \dfrac{1}{x} }$

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)