This question originates from Pinter's Abstract Algebra, Chapter 27, E1.
Recall the definition of $F(a)$. It is a field such that (i) $F \subseteq F(a)$; (ii) $a \in F(a)$; (iii) any field containing $F$ and $a$ contains $F(a)$. Use this definition to prove, where $F \subseteq K, c \in F$, and $a \in K$:
$\qquad F(a) = F(a + c)$ and $F(a) = F(ca)$. (Assume $c \ne 0$.)
First attempt:
Note $c$ is algebraic over $F$ if it is the root of some nonzero polynomial in $F[x]$. Let $K=F(a)=F[x]/\langle p(x)\rangle$ where $p(x)$ is a monic irreducible polynomial in $F[x]$ such that $a$ is a root. This implies $a$ is algebraic over $F$.
By the same reasoning of Exercise D1, if $a$ is algebraic over $F$, so are $a+c$ and $ca$ where $c \in F$ and $c\ne 0$.
Let $y=x-c$, so $a+c$ is a root of $p(y)$ and $F(a+c)=F[y]/\langle p(y)\rangle$. Let $w=x/c$, so $ca$ is a root of $p(w)$ and $F(ca)=F[w]/\langle p(w)\rangle$. Hence $F[x] / \langle p(x)\rangle = F(a) = F(a+c) = F(ca)$.
Second attempt:
\begin{align*} a\in F(a)&\implies a+c \in F(a) & \text{by closure of addition} \\ &\implies F(a+c)\subseteq F(a) & \text{by (iii)} \\ a+c\in F(a+c)&\implies a+c-c=a\in F(a+c) & \text{by closure of addition} \\ &\implies F(a)\subseteq F(a+c) & \text{by (iii)} \end{align*}
$\therefore F(a) = F(a+c)$.
Similarly,
\begin{align*} a\in F(a)&\implies ca \in F(a) & \text{by closure of multiplication} \\ &\implies F(ca)\subseteq F(a) & \text{by (iii)} \\ c\ne 0 \text{ and } ca\in F(ca)&\implies c^{-1}ca\in F(ca) & \text{by closure of multiplication} \\ &\implies F(a)\subseteq F(ca) & \text{by (iii)} \end{align*}
$\therefore F(a) = F(ca)$.
Does this look reasonable?
You don't need $a$ to be algebraic over $F$.
Note that by the first property $F$ is a subfield of both $F(a)$ and $F(a+c)$. Therefore, $c$ is an element of both of these fields. Hence, by closure of abelian group, we have, $$a+c\in F(a)\hspace{1cm}\text{and}\hspace{1cm}a+c-c\in F(a+c)\\ \implies a+c\in F(a)\hspace{1cm}\text{and}\hspace{1cm}a\in F(a+c)$$ Hence, by the third property, $F(a+c)$, being a field containing both $F$ and $a$, is a superfield of $F(a)$ and vice versa. Hence, the result.
The case of multiplication is exactly similar, you just have to use the closure of multiplication operation of the underlying ring instead of addition operation.