
solv que.
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solution:
_____________________________________________________________
Given:

_____________________________________________________________
we know that,
a² - b² = (a+b) (a-b)
and 1 power anything is 1,
so,
x² - 1 = (x+1)(x-1)
By using
a³ + b³ = (a+b) (a² - ab +b²)
a³ - b³ = (a-b) (a² + ab +b²)
(x³+1) =(x+1)(x²- x + 1)
& (x³-1) = (x-1) (x²+x+1)
so,
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=>![\frac{x^6 - 1}{x^2 - 1} = [(x^2 + 1) -x] [(x^2+1)+x] \frac{x^6 - 1}{x^2 - 1} = [(x^2 + 1) -x] [(x^2+1)+x]](https://tex.z-dn.net/?f=+%5Cfrac%7Bx%5E6+-+1%7D%7Bx%5E2+-+1%7D+%3D++%5B%28x%5E2+%2B+1%29+-x%5D+%5B%28x%5E2%2B1%29%2Bx%5D)
=>
=>
=>
__________________________________________________________
Hope it Helps!!
_____________________________________________________________
Given:
_____________________________________________________________
we know that,
a² - b² = (a+b) (a-b)
and 1 power anything is 1,
so,
x² - 1 = (x+1)(x-1)
By using
a³ + b³ = (a+b) (a² - ab +b²)
a³ - b³ = (a-b) (a² + ab +b²)
(x³+1) =(x+1)(x²- x + 1)
& (x³-1) = (x-1) (x²+x+1)
so,
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__________________________________________________________
Hope it Helps!!
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