Question

Factor completely 81x^8-1

Answer 1

The complete factorized form for the given expression is $\left(9 x^{4}+1\right)\left(3 x^{2}+1\right)\left(3 x^{2}-1\right)$

Step-by-step explanation:

Step 1: Given expression:

              $81 x^{8}-1$

Step 2: Trying to factor as a Difference of Squares

Factoring $81 x^{8}-1$

As we know the theory that the difference of two perfect squares, $A^{2}-B^{2}$  can be factored into (A+B) (A-B)

from this, when analysing, 81 is the square of 9, $x^{8}$ is the square of $x^{4}$. Hence, we can write the given expression as,

            $\left(9 x^{4}\right)^{2}-1^{2}$

By using the theory, we get

           $\left(9 x^{4}+1\right)\left(9 x^{4}-1\right)$

Again, we can further factorise the term $\left(9 x^{4}-1\right)$

$9 x^{4}$ is the square of $3 x^{2}$. Therefore, it can be expressed as below

           $\left(3 x^{2}+1\right)\left(3 x^{2}-1\right)$

Now, we can not factorise further the term $\left(3 x^{2}-1\right)$. Because it will come as $\sqrt{3} x$ (3 is not a square term). Thereby concluding that the complete factorisation for the given expression is $\left(9 x^{4}+1\right)\left(3 x^{2}+1\right)\left(3 x^{2}-1\right)$