Fully factorise
16x^36 -1
16x^36 - 1
= (4x^18)^2 - 1^2
= (4x^18 - 1)(4x^18 + 1)
= ((2x^9)^2 - 1^2)(4x^18 + 1)
= (2x^9 - 1)(2x^9 + 1)(4x^18 + 1)
To fully factorize the expression 16x^36 - 1, we can start by recognizing it as a difference of squares. The difference of squares formula is given by a^2 - b^2 = (a + b)(a - b).
In this case, we have (4x^18)^2 - 1^2. Applying the difference of squares formula, we get (4x^18 + 1)(4x^18 - 1).
Now, let's look at each of these factors separately.
Factor 1: 4x^18 + 1
This expression cannot be factored further using real numbers. Therefore, it remains as it is.
Factor 2: 4x^18 - 1
To factor this further, we recognize it as a difference of squares again. The expression 4x^18 - 1 can be written as (2x^9)^2 - 1^2. Applying the difference of squares formula, we get (2x^9 + 1)(2x^9 - 1).
Now, the fully factorized expression is:
16x^36 - 1 = (4x^18 + 1)(2x^9 + 1)(2x^9 - 1)