Factorisation of a2-1-a2-27 is equal to

The correct option is A (a 1a 5)(a 1a+5)The given expression is nbsp;a^2+1/a^2 27 This can be written as nbsp; a^2+1/a^2 2 25 = nbsp;a^2+1/a^2 2(a)(1/a) ...

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Byju's Answer

Standard IX

Mathematics

Difference of Two Squares

factorisation...

Question

factorisation of $a^{2} + \frac{1}{a^{2}} - 27$ is equal to .

(a - \frac{1}{a} - 5) (a - \frac{1}{a} + 5)

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(a - \frac{1}{a} - 5) (a - \frac{1}{a} - 5)

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(a - \frac{1}{a} + 5) (a - \frac{1}{a} + 5)

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(a + \frac{1}{a} - 5) (a + \frac{1}{a} + 5)

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Solution

The correct option is A $(a - \frac{1}{a} - 5) (a - \frac{1}{a} + 5)$
The given expression is $a^{2} + \frac{1}{a^{2}} - 27$
This can be written as
$a^{2} + \frac{1}{a^{2}} - 2 - 25$
= $a^{2} + \frac{1}{a^{2}} - 2 (a) (\frac{1}{a}) - 25$
= $(a - \frac{1}{a})^{2} - 5^{2}$
= $[(a - \frac{1}{a}) - 5] [(a - \frac{1}{a}) + 5]$
= $(a - \frac{1}{a} - 5) (a - \frac{1}{a} + 5)$

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factorisation of a2+1a2−27 is equal to .

The correct option is A (a−1a−5)(a−1a+5)The given expression is a2+1a2−27 This can be written as a2+1a2−2−25 = a2+1a2−2(a)(1a)−25 = (a−1a)2−52 = [(a−1a)−5] [(a−1a)+5] = (a−1a−5) (a−1a+5)

factorisation of $a^{2} + \frac{1}{a^{2}} - 27$ is equal to .