Question

$( {x}^{2} + \frac{1}{x {}^{2} } ) - 4(x + \frac{1}{x} ) + 6$
factorise it please ​

Answer 1

Answer:

$( {x}^{2} + \frac{1}{ {x}^{2} } ) - 4(x + \frac{1}{x}) + 6 = >$

$( {x}^{2} + \frac{1}{ {x}^{2} } + 2 \times x \times \frac{1}{x}) - 4(x + \frac{1}{x} )$

$+ 4 \: = >$

${(x + \frac{1}{x}) }^{2} - 4(x + \frac{1}{x}) + 4 \: = >$

Let

$x + \frac{1}{x} \: be \: y$

Then the above equation becomes

${y}^{2} - 2 \times 2 \times y + {2}^{2} = >$

${(y - 2)}^{2}$

Now we can substitute

$y = x + \frac{1}{x}$

So it can be factorised as

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

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