Math, asked by mohitverma2642005, 1 year ago

factorise p^8 -1/p^8

Answers

Answered by bibhanshuthapli
0
p^8 - 1/p^8

(p^4)^2 - (1/p^4)^2
(p^4 + 1/p^4) (p^4 - 1/p^4)
(p^4 + 1/p^4) ((p^2)^2 - (1/p^2)^2)
(p^4 + 1/p^4) (p^2 + 1/p^2)(p^2 - 1/p^2)
(p^4 + 1/p^4) (p^2 + 1/p^2)((p^√2)^2 - p^√2)^2)
(p^4 + 1/p^4) (p^2 + 1/p^2)(p+√2)(p-√2)
Answered by BhawnaAggarwalBT
1
here is your answer

{p}^{8}  - ( \frac{1}{p}  {)}^{8}  \\  \\  { ({p}^{4}) }^{2}  - ( \frac{1}{p {}}^{4}  {)}^{2}  \\  \\(  {p}^{4}  -   { \frac{1}{p} }^{4} )( {p}^{4} +  \frac{1}{p {}}^{4}  ) \\  \\(  {p}^{2}  -  { \frac{1}{p} }^{2} )( {p}^{2}  +  { \frac{1}{p} }^{2} )( {p}^{4}  +  { \frac{1}{p} }^{4} ) \\  \\ (p +  \frac{1}{p} )(p -  \frac{1}{p} )( {p}^{2}  +  { \frac{1}{p} }^{2} )( {p}^{4}  +  { \frac{1}{p} }^{4} )
hope his helps you
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