This is a problem from Algebra, Hungerford. Exercise V.5.21.

(a) Let $L$ and $M$ be intermediate fields of the extension $K \subset F$, of finite
dimension over $K$. Assume that $[LM : K:] = [L : K][M : K]$ and prove that $L \cap M = K$.

(b) The converse of (a) holds if $[L : K]$ or $[M : K]$ is 2.

(c) Using a real and a nonreal cube root of 2 give an example where $L \cap M = K$, $[L : K] = [M : K] = 3$, but $[LM : K] < 9$.

I've solved a) as follows:

$[L:K][L:M]=^{hyp}[LM:K]=[LM:L\cap M][L\cap M:K]\leq$ $[L:L\cap M][M:L\cap M][L\cap M:K]=[L:L\cap M][M:K]$.

So we have $[L:K]\leq [L:L\cap M]$ hence $L\cap M \subset K$.On the other hand, $K\subset L\cap M$ is obvious so we have the equality.

My problems come with b), could you help me? (c) is easy).

Assuming $car(k)\neq 2$, here is possible solution for b) Wlog, [L:K]=2. Thus $L/K$ is a Galois extension.
It is a well known result that (translation by $M$) $ML/M$ is a Galois extension and $$Gal(LM/M) \cong Gal(L/(L\cap M)).$$
Since $L\cap M=K$, we get $[LM:M]=[L:K]$, and then $$[LM:K]=[LM:M][M:K]=[L:K][M:K]$$