Let's calculate as simply as possible the value of

$$
a^{3}+b^{3}+3\left(a^{3} b+a b^{3}\right)+6\left(a^{3} b^{2}+a^{2} b^{3}\right)
$$

if $a+b=1$.
The expression $K$ can be transformed in various ways. We will present a few methods.

I. Solution. Using known identities, factoring, and then immediately writing 1 instead of the factor $a+b$, the value of the given expression $K$ is found to be 1:

$$
\begin{gathered}
K=(a+b)\left(a^{2}-a b+b^{2}\right)+3 a b\left(a^{2}+b^{2}\right)+6 a^{2} b^{2}(a+b)=a^{2}-a b+b^{2}+ \\
+3 a b\left[(a+b)^{2}-2 a b\right]+6 a^{2} b^{2}=(a+b)^{2}-3 a b+3 a b(1-2 a b)+6 a^{2} b^{2}=1
\end{gathered}
$$

Margit Dudás (Szeged, Tömörkény I. High School I. Class)

II. Solution. The three parts of $K$ are of the 3rd, 4th, and 5th degree, respectively, but based on the assumption, it can be transformed into a 3rd-degree expression. Indeed, $a^{2}+2 a b+b^{2}=(a+b)^{2}=a+b$, so

$$
\begin{gathered}
K=\left(a^{3}+b^{3}\right)+3 a b\left[a^{2}+b^{2}+2 a b(a+b)\right]=\left(a^{3}+b^{3}\right)+3 a b\left(a^{2}+b^{2}+\right. \\
+2 a b)=a^{3}+b^{3}+3 a b(a+b)=(a+b)^{3}=1
\end{gathered}
$$

Katalin Böszörményi (Budapest, Szilágyi E. High School I. Class)

III. Solution. By multiplying the first part by $1=a+b$, $K$ can also be written as a 4th-degree expression, and we can recognize the terms of the expansion of $(a+b)^{4}$:

$$
\begin{gathered}
K=(a+b)\left(a^{3}+b^{3}\right)+3 a^{3} b+3 a b^{3}+6 a^{2} b^{2}(a+b)=a^{4}+b^{4}+4 a^{3} b+4 a b^{3}+ \\
+6 a^{2} b^{2}=(a+b)^{4}=1^{4}=1
\end{gathered}
$$

Miklós Csúrrös (Nagykórös, Arany J. High School II. Class)