Question

factorize :- p⁴-81. ​

Answer 1

p⁴-81

{Identity to use: a²-b²= (a+b)(a-b)}

(p²)² - (9)²

(p²-9)(p²+9)

{the identity can again be used}

{(p)²-(3)²}(p²+9)

(p-3)(p+3)(p²+9)

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Answer 2

Answer:

${\red{ \bold{question \: = >}}} \\ \\ \sf{factorise \: {p}^{4} - 81} \\ \\ { \blue {\bold{solution \: = > }}} \\ \\ \sf{ we \: have \: \: \: }{p}^{4} - 81 = > { ({p}^{2} )}^{2} - ({9})^{2} \\ \\ \sf {now \: using \: {a}^{2} - {b}^{2} = (a + b)(a - b) \: we \: have} \\ \\ \sf{ {( {p}^{2} )}^{2} - ({9)}^{2} =( {p}^{2} + 9)( {p}^{2} - 9)} \\ \\ \sf{we \: can \: factorise \: {p}^{2} - 9 \: further \: as} \\ \\ \sf{ {p}^{2} - 9 = ({p})^{2} - ({3})^{2} } \\ \\ \sf {= (p + 3)(p - 3)} \\ \\ { \boxed{ \bold{\sf{ \therefore \: {p}^{4} - 81 = (p + 3)(p - 3)( {p}^{2} + 9)}}}}$

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