To expand the expression $(2 + x)^3$, we use the binomial theorem, which states:

$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

For $(2 + x)^3$, we have $a = 2$, $b = x$, and $n = 3$. Expanding it gives:

$(2 + x)^3 = \binom{3}{0}2^3x^0 + \binom{3}{1}2^2x^1 + \binom{3}{2}2^1x^2 + \binom{3}{3}2^0x^3$

Simplifying each term:

$= 1 \cdot 8 \cdot 1 + 3 \cdot 4 \cdot x + 3 \cdot 2 \cdot x^2 + 1 \cdot 1 \cdot x^3$ $= 8 + 12x + 6x^2 + x^3$

So the expanded form of $(2 + x)^3$ is:

$(2 + x)^3 = x^3 + 6x^2 + 12x + 8$

Would you like more details or have any specific questions?

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Here are five related questions:

1. How would you factor the expression $x^3 + 6x^2 + 12x + 8$ back into $(2 + x)^3$?
2. Can you find the derivative of $x^3 + 6x^2 + 12x + 8$ with respect to $x$?
3. How would you evaluate $(2 + x)^3$ for a specific value of $x$, like $x = 1$?
4. What is the integral of $x^3 + 6x^2 + 12x + 8$ with respect to $x$?
5. How can you solve the equation $(2 + x)^3 = 27$ for $x$?

Tip: Expanding binomials using the binomial theorem is a powerful technique, especially for higher powers. It allows you to express a binomial raised to a power as a sum of terms, which can be useful in calculus and algebra.