The problem is to factor the expression $3x^6 + 81y^6$.

1. Problem Description

The problem is to factor the expression $3x^6 + 81y^6$.

2. Solution Steps

First, we can factor out a 3 from both terms:

$3x^6 + 81y^6 = 3(x^6 + 27y^6)$

Notice that $x^6 = (x^2)^3$ and $27y^6 = (3y^2)^3$. We can use the sum of cubes formula:

$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$

In this case, $a = x^2$ and $b = 3y^2$. Substituting these values into the formula, we have:

$x^6 + 27y^6 = (x^2)^3 + (3y^2)^3 = (x^2 + 3y^2)((x^2)^2 - (x^2)(3y^2) + (3y^2)^2)$

$x^6 + 27y^6 = (x^2 + 3y^2)(x^4 - 3x^2y^2 + 9y^4)$

Substituting this back into the original expression:

$3(x^6 + 27y^6) = 3(x^2 + 3y^2)(x^4 - 3x^2y^2 + 9y^4)$

3. Final Answer

The factored expression is $3(x^2 + 3y^2)(x^4 - 3x^2y^2 + 9y^4)$.