Question

question
factor completely.
$81x^{4}y^{6}-1$
answer attempt 1 out of 10

Explanation:

Step1: Recognize difference of squares

$81x^4y^6 - 1 = (9x^2y^3)^2 - (1)^2$

Step2: Apply difference of squares formula

$(9x^2y^3 - 1)(9x^2y^3 + 1)$

Step3: Factor first term again

$9x^2y^3 - 1 = (3xy^{3/2})^2 - 1^2$, rewrite as $(3xy\sqrt{y} - 1)(3xy\sqrt{y} + 1)$

Step4: Combine all factors

$(3xy\sqrt{y} - 1)(3xy\sqrt{y} + 1)(9x^2y^3 + 1)$
Or, using integer exponents:

Step1: Recognize difference of squares

$81x^4y^6 - 1 = (9x^2y^3)^2 - 1^2$

Step2: Factor first difference of squares

$(9x^2y^3 - 1)(9x^2y^3 + 1)$

Step3: Factor $9x^2y^3 -1$ as difference of squares (rational exponents)

$9x^2y^3 -1 = (3xy^{3/2})^2 -1^2 = (3xy^{3/2}-1)(3xy^{3/2}+1)$

Step4: Rewrite with integer exponents (alternative)

$81x^4y^6 -1=(9x^2y^3 -1)(9x^2y^3 +1)$
If restricting to integer exponents, this is fully factored over integers; if allowing rational exponents, it factors further to $(3xy\sqrt{y}-1)(3xy\sqrt{y}+1)(9x^2y^3+1)$

Answer:

$(3xy\sqrt{y} - 1)(3xy\sqrt{y} + 1)(9x^2y^3 + 1)$
Or, factored over integers: $(9x^2y^3 - 1)(9x^2y^3 + 1)$