The correct option is A (p^2+9)(p^2 9) #160; p^4 81 =p^4 (3)^4 = (p^2)^2 (3^2)^2 Using, [a^2 b^2=(a+b)(a b)] =(p^2+3^2)(p^2 3^ ...

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Expansion of (x^2 - y^2)

Factorise: p...

Question

Factorise:
$p^{4} - 81$

(p^{2} + 9) (p^{2} - 9)

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(p^{4} - 27^{4})

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(p^{3} - 3^{3})

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(p^{4} - 3)

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Solution

The correct option is A $(p^{2} + 9) (p^{2} - 9)$
$p^{4} - 81$
$= p^{4} - (3)^{4}$
$= (p^{2})^{2} - (3^{2})^{2}$ Using, $[a^{2} - b^{2} = (a + b) (a - b)]$
$= (p^{2} + 3^{2}) (p^{2} - 3^{2})$ Using, $[a^{2} - b^{2} = (a + b) (a - b)]$
$= (p^{2} + 9) (p^{2} - 9)$

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