Answer:
(3x^2 - 1)(3x^2 + 1)(9x^4 + 1).
Step-by-step explanation:
Using the identity for the difference of 2 squares;
a^2 - b^2 = (a - b)(a + b)
we put a^2 = 81x^8 and b^2 = 1 giving
a = 9x^4 and b = 1, so:
81x^8 − 1 = (9x^4 - 1)(9x^4 + 1)
Applying the difference of 2 squares to 9x^4 - 1:
= (3x^2 - 1)(3x^2 + 1)(9x^4 + 1).
The required factor for [tex]81x^8 -1.[/tex] is [tex](3x^2 -1)(3x^2 + 1)(9x^4 + 1)[/tex].
To Factorize [tex]81x^8 -1[/tex].
factors can be defined splitting the value into multipliable values.
To Factorize [tex]81x^8 -1[/tex]
[tex]=(9x4)^2-1[/tex]
Now using identity [tex]a^2-b^2=(a+b)(a-b)[/tex]
here [tex]a = 9x^4, b =1[/tex]
[tex]=(9x4)^2-1\\=(9x^4-1)(9x^4+1)\\=[(3x^2)^2-1](9x^4+1)\\[/tex]
One more time using identity [tex]a^2-b^2=(a+b)(a-b)[/tex]
Where a = 3x² and b = 1
[tex]=[(3x^2)^2-1](9x^4+1)\\=(3x^2-1)(3x^2+1)(9x^4+1)[/tex]
Required factors of [tex]81x^8 -1[/tex] is [tex](3x^2 -1)(3x^2 + 1)(9x^4 + 1)[/tex]
Thus, the required factor for [tex]81x^8 -1.[/tex] is [tex](3x^2 -1)(3x^2 + 1)(9x^4 + 1)[/tex].
Learn more about factors here:
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