Question

factorise x^2 + 1/x^2 - 2 -3x + 3/x

Answer 1

Answer:

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

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

Method:-

First we split the expression in the following way:-

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

In the first part, we will use Identity 2, which states:-

$(a - b)^{2} = a^{2} + b^{2} - 2ab$

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

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

In the second part, we can take out 3 as common and we will get:-

$-[3x - \frac{3}{x}] = -3[x - \frac{1}{x}]$

We get the term:-

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

Verify:-

When we solve the factored term, the answer should be equal to the non - factored term.

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

LHS = RHS

Hence, verified

Hope it helps

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