The expansion of $(x + y)^6$ is done using the binomial theorem, which states:

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

For $(x + y)^6$, $a = x$, $b = y$, and $n = 6$. The expansion is:

$(x + y)^6 = \sum_{k=0}^6 \binom{6}{k} x^{6-k} y^k$

Calculating each term:

1. $\binom{6}{0} x^6 y^0 = 1x^6 = x^6$
2. $\binom{6}{1} x^5 y^1 = 6x^5y$
3. $\binom{6}{2} x^4 y^2 = 15x^4y^2$
4. $\binom{6}{3} x^3 y^3 = 20x^3y^3$
5. $\binom{6}{4} x^2 y^4 = 15x^2y^4$
6. $\binom{6}{5} x^1 y^5 = 6xy^5$
7. $\binom{6}{6} x^0 y^6 = y^6$

Thus, the expansion is:

$x^6 + 6x^5y + 15x^4y^2 + 20x^3y^3 + 15x^2y^4 + 6xy^5 + y^6$

Would you like more details or assistance with related problems?

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Related Questions:

1. Can you expand $(x - y)^6$ using the same method?
2. How do you calculate the binomial coefficient $\binom{6}{k}$?
3. What is the general term in the expansion of $(x + y)^n$?
4. How does the binomial theorem relate to Pascal's Triangle?
5. What happens when $x = 1$ in the expansion of $(x + y)^n$?

Tip:

When expanding a binomial using the binomial theorem, ensure coefficients are calculated correctly using (\binom{n}{k} = \