Question

Question -

Expand -

\sf{ ( p - \dfrac{1}{p} ) ( p + \dfrac{1}{p} ) ( p^2 + \dfrac{1}{ p^2 } )( p^4 + \dfrac{1}{ p^4 } ) }(p−p1​)(p+p1​)(p2+p21​)(p4+p41​)

Solution -

\begin{lgathered}\sf{ ( p - \dfrac{1}{p} ) ( p + \dfrac{1}{p} ) ( p^2 + \dfrac{1}{ p^2 } )( p^4 + \dfrac{1}{ p^4 } ) } \\ \\ \sf{ => ( p^2 - \dfrac{1}{ p^2 } )( p^2 + \dfrac{1}{ p^2 } )( p^4 + \dfrac{1}{ p^4 } ) } \\ \\ \sf{ => ( p^4 - \dfrac{1}{ p^4} )( p^4 + \dfrac{1}{ p^4 } ) } \\ \\ \sf{ => (p^8 - \dfrac{1}{ p^8 } ) } \\ \\ \sf{ \bold{ Hence \: Expanded } }\end{lgathered}(p−p1​)(p+p1​)(p2+p21​)(p4+p41​)=>(p2−p21​)(p2+p21​)(p4+p41​)=>(p4−p41​)(p4+p41​)=>(p8−p81​)HenceExpanded​

Additional Identities -

\begin{lgathered}\sf{ (a - b )( a + b ) = a^2 - b^2 } \\ \\ \sf{ ( a + b )^2 = a^2 + 2ab + b^2 } \\ \\ \sf{ ( a - b )^2 = a^2 - 2ab + b^2 }\end{lgathered}(a−b)(a+b)=a2−b2(a+b)2=a2+2ab+b2(a−b)2=a2−2ab+b2​

prove it
​

Answer 1

Answer:

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Answer 2

Answer:

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