Question

Factorise: x^2 + 1/x^2 - 6

Answer 1

Answer:

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Answer 2

Answer:

$\underline{ \bf{\dag}\:\:\large{\textit{Method 1 :}}}$

$:\implies\tt x^2+\dfrac{1}{x^2}-6\\\\\\:\implies\tt x^2 + \dfrac{1}{x^2} + 2 - 8\\\\\\:\implies\tt (x)^2 + \bigg( \dfrac{1}{x} \bigg)^2 + \bigg(2 \times x \times \dfrac{1}{x} \bigg) - (2 \sqrt{2})^2\\\\\\:\implies\tt \bigg(x + \dfrac{1}{x}\bigg)^2 -(2 \sqrt{2})^2\\\\\\:\implies\underline{\boxed{\tt \bigg(x + \dfrac{1}{x} + 2 \sqrt{2} \bigg) \bigg(x + \dfrac{1}{x} -2 \sqrt{2} \bigg)}}$

$\rule{200}{1}$

$\underline{ \bf{\dag}\:\:\large{\textit{Method 2 :}}}$

$:\implies\tt x^2+\dfrac{1}{x^2}-6\\\\\\:\implies\tt x^2 + \dfrac{1}{x^2} - 2 - 4\\\\\\:\implies\tt (x)^2 + \bigg( \dfrac{1}{x} \bigg)^2 - \bigg(2 \times x \times \dfrac{1}{x} \bigg) - (2)^2\\\\\\:\implies\tt \bigg(x - \dfrac{1}{x}\bigg)^2 -(2)^2\\\\\\:\implies\underline{\boxed{\tt \bigg(x - \dfrac{1}{x} + 2\bigg) \bigg(x - \dfrac{1}{x} -2\bigg)}}$