To be calculated

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

value, if $a+b=1$.
I. solution: Let's denote our expression by $N$. Multiply the sum of the first two terms by $(a+b)=1$:

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

Pak To Ha (Pécs, Bányaip. techn. II. o. t.)

II. solution: Always keeping in mind that $a+b=1$,

$$
\begin{aligned}
N & =a^{3}+b^{3}+3 a b\left(a^{2}+b^{2}\right)+6 a^{2} b^{2}(a+b)= \\
& =a^{3}+b^{3}+3 a b\left(a^{2}+b^{2}\right)+6 a^{2} b^{2}=a^{3}+b^{3}+3 a b\left(a^{2}+2 a b+b^{2}\right)= \\
& =a^{3}+b^{3}+3 a b(a+b)^{2}=a^{3}+b^{3}+3 a b(a+b)= \\
& =a^{3}+3 a^{2} b+3 a b^{2}+b^{3}=(a+b)^{3}=1
\end{aligned}
$$

Tóth László (Miskolc, Vill. energia ip. techn. II. o)

III. solution: Keeping in mind that $a+b=1$,

$$
\begin{aligned}
N & =a^{3}+b^{3}+3 a^{3} b+3 a b^{3}+6 a^{2} b^{2}(a+b)= \\
& =a^{3}+b^{3}+3 a^{3} b+3 a b^{3}+3 a^{2} b^{2}+3 a^{2} b^{2}= \\
& =a^{3}+b^{3}+3 a^{2} b(a+b)+3 a b^{2}(a+b)=(a+b)^{3}=1
\end{aligned}
$$

Cserteg István (Bp. VIII., Széchenyi g. II. o. t.)

IV. solution: Our expression can also be written as

$$
\begin{aligned}
N & =(a+b)\left(a^{2}-a b+b^{2}\right)+3 a b\left(a^{2}+b^{2}\right)+6 a^{2} b^{2}(a+b)= \\
& =a^{2}-a b+b^{2}+3 a b\left(a^{2}+b^{2}\right)+6 a^{2} b^{2}= \\
& =a^{2}-a b+b^{2}+3 a b\left(a^{2}+2 a b+b^{2}\right)= \\
& =a^{2}-a b+b^{2}+3 a b(a+b)^{2}=a^{2}-a b+b^{2}+3 a b= \\
& =a^{2}+2 a b+b^{2}=(a+b)^{2}=1
\end{aligned}
$$

Décsey Julianna (Karcag, Gábor Áron g. II. o. t.)