Show that $\alpha(x+y) = \alpha(x)α(y) + \beta(x)\gamma(y) + \beta(y)\gamma(x)$ for every $x, y \in\mathbb R$. $$\alpha(x) =\sum_{j=0}^\infty \frac{x^{3j}}{(3j)!}$$
$$\beta(x) = \sum_{j=0}^\infty \frac{x^{3j+2}}{(3j+2)!}$$
$$\gamma(x) = \sum_{j=0}^\infty \frac{x^{3j+1}}{(3j+1)!}$$
Show that $\alpha(x+y) = \alpha(x)α(y) + \beta(x)\gamma(y) + \beta(y)\gamma(x)$ for every $x, y \in\mathbb R$.
Any help will be appreciated. 
 A: Hint:
Let's assume that we have 
$\sigma=e^{i 2\pi /3}=-1/2+i\sqrt{3}/2$
$\sigma^3-1=0$
$$(\sigma-1)(\sigma^2+\sigma+1)=0$$
$\sigma^2+\sigma+1=0$
$$e^{ x} =\sum_{j=0}^\infty \frac{ x^{ j}}{(j)!}=1+\frac{ x^{1}}{1!}+\frac{ x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{  x^{4}}{4!}+...$$
$$e^{\sigma x} =\sum_{j=0}^\infty \frac{\sigma^{j} x^{ j}}{(j)!}=1+\frac{\sigma x^{1}}{1!}+\frac{\sigma ^2 x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{\sigma  x^{4}}{4!}...$$
$$ e^{\sigma^2 x} =\sum_{j=0}^\infty \frac{\sigma^{2j} x^{ j}}{(j)!}=1+\frac{\sigma^2 x^{1}}{1!}+\frac{\sigma x^{2}}{2!}+\frac{ x^{3}}{3!}+\frac{\sigma^2  x^{4}}{4!}+...$$
$$ e^{ x} +e^{\sigma x}+e^{\sigma^2 x} =3+\frac{(1+\sigma+\sigma^2) x^{1}}{1!}+\frac{(1+\sigma+\sigma^2) x^{2}}{2!}+\frac{3 x^{3}}{3!}+\frac{(1+\sigma+\sigma^2)  x^{4}}{4!}+...$$
$$ e^{ x} +e^{\sigma x}+e^{\sigma^2 x} =3(1+\frac{ x^{3}}{3!}+\frac{ x^{6}}{6!}+\frac{ x^{9}}{9!}+.....)$$
$$\alpha(x)=\frac{ e^{ x} +e^{\sigma x}+e^{\sigma^2 x}}{3} $$
And you know that
$$\beta(x)=\frac{ x^{2}}{2!}+\frac{ x^{5}}{5!}+\frac{ x^{8}}{8!}+.....=\alpha'(x)=\frac{ e^{ x} +\sigma e^{\sigma x}+\sigma^2 e^{\sigma^2 x}}{3}$$
$$\beta(x)=\alpha'(x)=\frac{ e^{ x} +\sigma e^{\sigma x}+\sigma^2 e^{\sigma^2 x}}{3}$$
$$\gamma(x)=\frac{ x^{1}}{1!}+\frac{ x^{4}}{4!}+\frac{ x^{7}}{7!}+.....=\alpha''(x)=\frac{ e^{ x} +\sigma^2 e^{\sigma x}+\sigma e^{\sigma^2 x}}{3}$$
$$\gamma(x)=\alpha''(x)=\frac{ e^{ x} +\sigma^2 e^{\sigma x}+\sigma e^{\sigma^2 x}}{3}$$
Now you can use them prove your identity
$$\alpha(x)\alpha(y) + \beta(x)\gamma(y) + \beta(y)\gamma(x)=$$
$$(\frac{ e^{ x} +e^{\sigma x}+e^{\sigma^2 x}}{3})(\frac{ e^{ y} +e^{\sigma y}+e^{\sigma^2 y}}{3})+(\frac{ e^{ x} +\sigma e^{\sigma x}+\sigma^2 e^{\sigma^2 x}}{3})(\frac{ e^{ y} +\sigma^2 e^{\sigma y}+\sigma e^{\sigma^2 y}}{3})+(\frac{ e^{ y} +\sigma e^{\sigma y}+\sigma^2 e^{\sigma^2 y}}{3})(\frac{ e^{ x} +\sigma^2 e^{\sigma x}+\sigma e^{\sigma^2 x}}{3})  $$
Can you continue after that??
A: $$\alpha(x+y) \\ =\sum_{j=0}^\infty \frac{(x+y)^{3j}}{(3j)!}\\=\sum_{j=0}^\infty\sum_{k=0}^{3j} \binom {3j} k \frac{x^ky^{3j-k}}{(3j)!}\\
=\sum_{j=0}^\infty\sum_{k=0}^{3j} \frac {(3j)!}{k!(3j-k)!} \frac{x^ky^{3j-k}}{(3j)!}\\
=\sum_{j=0}^\infty\sum_{k=0}^{3j} \frac{x^ky^{3j-k}}{k!(3j-k)!}\\
=\sum_{j,k, k\le 3j} \frac{x^{k}}{k!} \frac{y^{3j-k}}{(3j-k)!}$$
Let $i=3j-k$.
$$\sum_{j,k, k\le 3j} \frac{x^{k}}{k!} \frac{y^{3j-k}}{(3j-k)!}\\
=\sum_{3\mid i+k} \frac{x^{k}}{k!} \frac{y^{i}}{i!}$$
Let $n$ be integer part of $\frac k 3$, one of these hold:
$k=3n, k=3n+2, k=3n+1$
$$\sum_{3\mid i+k} \frac{x^{k}}{k!} \frac{y^{i}}{i!}\\
=\sum_{n, 3\mid i} \frac{x^{3n}}{(3n)!} \frac{y^{i}}{i!}+
\sum_{n, 3\mid i+1} \frac{x^{3n+2}}{(3n+2)!} \frac{y^{i}}{i!}+
\sum_{n, 3\mid i+2} \frac{x^{3n+1}}{(3n+1)!} \frac{y^{i}}{i!}$$
Let $m$ be integer part of $\frac i 3$.
$$\sum_{n, 3\mid i} \frac{x^{3n}}{(3n)!} \frac{y^{i}}{i!}+
\sum_{n, 3\mid i+1} \frac{x^{3n+2}}{(3n+2)!} \frac{y^{i}}{i!}+
\sum_{n, 3\mid i+2} \frac{x^{3n+1}}{(3n+1)!} \frac{y^{i}}{i!}\\
=\sum_{n, m} \frac{x^{3n}}{(3n)!} \frac{y^{3m}}{(3m)!}+
\sum_{n, m} \frac{x^{3n+2}}{(3n+2)!} \frac{y^{3m+1}}{(3m+1)!}+
\sum_{n, m} \frac{x^{3n+1}}{(3n+1)!} \frac{y^{3m+2}}{(3m+2)!}\\
=\alpha(x)\alpha(y)+\beta(x)\gamma(y)+\gamma(x)\beta(y)$$.
