The correct option is B ( x+ 1 x +1 ) ( x ^ 2 + 1 x ^ 2 1 x x ) x^3+1x^3 2 x^3+1x^3+1 3 ∵ x^3+y^3+z^3 3xyz= ( x+y+z ) ( x^2+y^2+z ...

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Difference of Two Squares

Factorise: ...

Question

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

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

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

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

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

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Solution

The correct option is B $(x + \frac{1}{x} + 1) (x^{2} + \frac{1}{x^{2}} - \frac{1}{x} - x)$
$x^{3} + \frac{1}{x^{3}} - 2$
$x^{3} + \frac{1}{x^{3}} + 1 - 3$
$∵ x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x)$
$\Rightarrow x^{3} + \frac{1}{x^{3}} + 1^{3} - 3 \times x \times \frac{1}{x} \times 1$
$= (x + \frac{1}{x} + 1) (x^{2} + \frac{1}{x^{2}} + 1 - 1 - \frac{1}{x} - x)$
$= (x + \frac{1}{x} + 1) (x^{2} + \frac{1}{x^{2}} - \frac{1}{x} - x)$

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