81y4-18y2+1

Rewrite y4 as (y2)2.

81(y2)2-18y2+1

Let u=y2. Substitute u for all occurrences of y2.

81u2-18u+1

Rewrite 81u2 as (9u)2.

(9u)2-18u+1

Rewrite 1 as 12.

(9u)2-18u+12

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

18u=2⋅(9u)⋅1

Rewrite the polynomial.

(9u)2-2⋅(9u)⋅1+12

Factor using the perfect square trinomial rule a2-2ab+b2=(a-b)2, where a=9u and b=1.

(9u-1)2

(9u-1)2

Replace all occurrences of u with y2.

(9y2-1)2

Rewrite 9y2 as (3y)2.

((3y)2-1)2

Rewrite 1 as 12.

((3y)2-12)2

Since both terms are perfect squares, factor using the difference of squares formula, a2-b2=(a+b)(a-b) where a=3y and b=1.

((3y+1)(3y-1))2

Apply the product rule to (3y+1)(3y-1).

(3y+1)2(3y-1)2

Factor 81y^4-18y^2+1