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

Math
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
Factor using the perfect square rule.
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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

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