Question

x²+1/x²-3 factories ​

Answer 1

Answer:

x^2+1/x^2-3

=(x-1/x)^2+2x.1/x-3

=(x-1/x)^2+2-3

=(x-1/x)^2-1

=(x-1/x)^2-(1)^2

=(x-1/x+1)(x-1/x-1)

Answer 2

Required Answer:-

Given To Factorise:

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

Solution:

We have,

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

This can be written as,

$\sf = {x}^{2} + \dfrac{1}{ {x}^{2} } - 2 - 1$

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

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

This is in (a - b)² form. Factorise it.

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

Now, a² - b² = (a + b)(a - b). Factorise using this identity,

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

This is the required factorised form for the question.

Answer:

• $\sf \bigg(x - \dfrac{1}{x} + 1\bigg) \bigg(x - \dfrac{1}{x} - 1 \bigg)$

Identity Used:

• (a - b)² = a² - 2ab + b²
• a² - b² = (a + b)(a - b)