Let's solve this step-by-step. The expression is:

$(x + y)^3 + 8y^3$

We recognize this as a sum of cubes. The general form of a sum of cubes is:

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

Here, $a = (x + y)$ and $b = 2y$, since $8y^3 = (2y)^3$. So, applying the formula:

$(x + y)^3 + (2y)^3 = \left((x + y) + 2y\right)\left((x + y)^2 - (x + y)(2y) + (2y)^2\right)$

Now simplify each part:

First part:

$(x + y) + 2y = x + 3y$

Second part:

$(x + y)^2 = x^2 + 2xy + y^2$

$(x + y)(2y) = 2xy + 2y^2$

$(2y)^2 = 4y^2$

Now substitute back into the expression:

$\left(x + 3y\right)\left(x^2 + 2xy + y^2 - (2xy + 2y^2) + 4y^2\right)$

Simplifying the terms inside the second bracket:

$x^2 + 2xy + y^2 - 2xy - 2y^2 + 4y^2 = x^2 + (y^2 - 2y^2 + 4y^2) = x^2 + 3y^2$

So, the factorized form is:

$(x + 3y)(x^2 + 3y^2)$

Final Answer:

$(x + 3y)(x^2 + 3y^2)$

Would you like more details on any of the steps? Here are 5 related questions for practice:

1. How would you factorize $x^3 + 27$?
2. What is the factored form of $a^3 - b^3$?
3. Can you apply the sum of cubes formula to $x^3 + 8$?
4. How would you expand $(x + y)^3$?
5. What is the general form for factoring a difference of cubes?

Tip: When factoring sums or differences of cubes, remember to recognize the cube terms and apply the correct formulas.