The given equation is:

$y = \sqrt[3]{x} + 2$

To solve for $x$, we need to isolate $x$ on one side of the equation. Here's how to proceed step by step:

1. Subtract 2 from both sides to move the constant term: $y - 2 = \sqrt[3]{x}$

2. Cube both sides to eliminate the cube root: $(y - 2)^3 = x$

Thus, the equation for $x$ in terms of $y$ is:

$x = (y - 2)^3$

Now, let's calculate some values of $x$ for integer values of $y$ (without decimals):

For $y = 0, 1, 2, 3, 4$:

• When $y = 0$:
  $x = (0 - 2)^3 = (-2)^3 = -8$
• When $y = 1$:
  $x = (1 - 2)^3 = (-1)^3 = -1$
• When $y = 2$:
  $x = (2 - 2)^3 = 0^3 = 0$
• When $y = 3$:
  $x = (3 - 2)^3 = 1^3 = 1$
• When $y = 4$:
  $x = (4 - 2)^3 = 2^3 = 8$

Therefore, the pairs of $x$ and $y$ values are:

• $y = 0 \Rightarrow x = -8$
• $y = 1 \Rightarrow x = -1$
• $y = 2 \Rightarrow x = 0$
• $y = 3 \Rightarrow x = 1$
• $y = 4 \Rightarrow x = 8$

Would you like more details or have any other questions?

Relative questions:

1. What would the solution for $x$ be if $y = -1$?
2. How does the behavior of the equation change for negative values of $y$?
3. Can you explain why cubing eliminates the cube root in this context?
4. What happens to the value of $x$ as $y$ increases significantly, say for $y = 10$?
5. How would you solve this equation if there were an additional term in the cube root expression?

Tip:

When solving equations involving roots, always consider inversing the root operation to simplify the equation (e.g., cube both sides for cube roots).