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Leona [35]
3 years ago
8

Factor completely 81x^8-1

Mathematics
1 answer:
marysya [2.9K]3 years ago
8 0

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)

<u>Step-by-step explanation:</u>

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)

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