Question

factorise X square + 1 / x square - 3

Answer 1

Answer:

$(x-{\frac{1}{x}}-1)(x+{\frac{1}{x}}+1)$

Step-by-step explanation:

The given equation is:

$x^2+{\frac{1}{x^2}-3$

which can be written as:

=$x^2+{\frac{1}{x^2}-2-1$

=$(x-{\frac{1}{x}})^2-1^2$

=$(x-{\frac{1}{x}}-1)(x+{\frac{1}{x}}+1)$

which is the required factorized form.

Answer 2

Answer:

$x^{2}+\frac{1}{x^{2}}-3=(x-\frac{1}{x}+1)(x-\frac{1}{x}-1)$

Step-by-step explanation:

$Given, \\x^{2}+\frac{1}{x^{2}}-3$

$=[x^{2}+\frac{1}{x^{2}}-2]-1$

$=[x^{2}+\frac{1}{x^{2}}-2\times x\times \frac{1}{x}]-1$

$=\left(x-\frac{1}{x}\right)^{2}-1^{2}$

/* By algebraic identity:

a²-2ab+b² = (a-b)² */

$=(x-\frac{1}{x}+1)(x-\frac{1}{x}-1)$

/* By algebraic identity:

a²-b² = (a+b)(a-b) */

Therefore,

$x^{2}+\frac{1}{x^{2}}-3=(x-\frac{1}{x}+1)(x-\frac{1}{x}-1)$

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