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)`

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
`√(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)`