Let $R$ be an $n$-dimensional Noetherian ring with proper ideal $I$. If $I = \mathfrak{a} \cap \mathfrak{b}$ and $H^n _\mathfrak{a}(M) = H^n _\mathfrak{b}(M) = 0$, for some $R$-module $M$, show $H^n_I(M) = 0$.
I ask this in reference to exercise 8.1.5 in Brodmann and Sharp's Local Cohomology book, where they ask a similar question instead using minimal primes belonging to $I$. So, if it is necessary, one may assume that $\mathfrak{a}$ and $\mathfrak{b}$ are minimal primes belonging to $I$.
Using Mayer–Vietoris, 3.2.3, and Vanishing theorem, 6.1.2, we have:
$$H^n _\mathfrak{a}(M) \oplus H^n _\mathfrak{b}(M) = 0\to H^n_I(M) \to H^{n+1}_{\mathfrak a+\mathfrak b}(M) = 0.$$