The complete factorized form for the given expression is [tex]\left(9 x^{4}+1\right)\left(3 x^{2}+1\right)\left(3 x^{2}-1\right)[/tex]
Step-by-step explanation:
Step 1: Given expression:
[tex]81 x^{8}-1[/tex]
Step 2: Trying to factor as a Difference of Squares
Factoring [tex]81 x^{8}-1[/tex]
As we know the theory that the difference of two perfect squares, [tex]A^{2}-B^{2}[/tex] can be factored into (A+B) (A-B)
from this, when analysing, 81 is the square of 9, [tex]x^{8}[/tex] is the square of [tex]x^{4}[/tex]. Hence, we can write the given expression as,
[tex]\left(9 x^{4}\right)^{2}-1^{2}[/tex]
By using the theory, we get
[tex]\left(9 x^{4}+1\right)\left(9 x^{4}-1\right)[/tex]
Again, we can further factorise the term [tex]\left(9 x^{4}-1\right)[/tex]
[tex]9 x^{4}[/tex] is the square of [tex]3 x^{2}[/tex]. Therefore, it can be expressed as below
[tex]\left(3 x^{2}+1\right)\left(3 x^{2}-1\right)[/tex]
Now, we can not factorise further the term [tex]\left(3 x^{2}-1\right)[/tex]. Because it will come as [tex]\sqrt{3} x[/tex] (3 is not a square term). Thereby concluding that the complete factorisation for the given expression is [tex]\left(9 x^{4}+1\right)\left(3 x^{2}+1\right)\left(3 x^{2}-1\right)[/tex]
