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प्रश्न
Factorise:
`"p"^2 + (1)/"p"^2 - 3`
उत्तर
`"p"^2 + (1)/"p"^2 - 3`
= `"P"^2 + (1)/"p"^2 - 2 - 1`
= `("p"^2 + (1)/"p"^2 - 2 xx "p" xx 1/"p") -1`
= `("p" - 1/"p")^2 - 1^2`
= `("p" - 1/"p" + 1)("p" - 1/"p" - 1)`.
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