### Instruction: In the equation $w^3+x^3+y^3=z^3$, $w^3$, $x^3$, $y^3$, and $z^3$ are distinct, consecutive positive perfect cubes listed in ascending order. What is the smallest possible value of $z$?### Response: To find the smallest possible value of $z$, I want to find the smallest possible values of $w$, $x$, and $y$ that satisfy the equation.
Since $w^3$ is the smallest perfect cube in the equation, it must be $1^3=1$, the smallest positive perfect cube.
That means $x^3$ must be the next perfect cube after $1^3$, which is $2^3=8$.
Similarly, $y^3$ must be the next perfect cube after $2^3$, which is $3^3=27$.
Now I have $w^3+x^3+y^3=1+8+27=36$, and I want to find $z^3$ such that $z^3=36$.
To do this, I can take the cube root of both sides of the equation and get $z=\sqrt[3]{36}$.
Since $z$ is a positive integer, I need to find the smallest integer that is greater than or equal to $\sqrt[3]{36}$.
I can estimate $\sqrt[3]{36}$ by trying some perfect cubes near 36.
For example, $3^3=27$ is too small and $4^3=64$ is too big, so $\sqrt[3]{36}$ must be between 3 and 4.
In fact, $\sqrt[3]{36}$ is closer to 4 than 3, since 36 is closer to 64 than 27.