We discuss on the field of complex numbers. Let $X$ be a smooth projective hypersurface of dimension $n \geq 4$ in $\mathbb{P}^{n+1}(\mathbb{C})$. Assume that for a general point $x \in X$, there exists a 4-dimensional linear subspace $L\_x$ in $\mathbb{P}^{n+1}$ passing through $x$ such that the set-theoretic intersection of $L\_x$ and $X$ is an irreducible cubic hypersurface in $L\_x$. Then does $X$ have to be a cubic hypersurface?


We know that if $L\_x$ is a general 4-dimensional linear subspace passing through $x$, then $L\_x \cap X$ is a hypersurface with the same degree as $X$. The trouble is that if $L\_x$ is special, then maybe the scheme-theoretic intersection of $L\_x$ and $X$ is reduced (for example, the restriction on $L\_x$ of the equation $f$ of $X$ in $\mathbb{P}^{n+1}$ maybe a square of a cubic equation and in this case $\deg(X)=6$). Hence, without the generality assumption of $x \in X$, the answer of my question is negative.


Now we have assumed the point $x$ is general in $X$ and that $X$ is smooth. Probably these assumptions can imply that $\deg(X)=3$. But I do not how to prove it.



Special linear sections of a hypersurface
The answer is "no" if the degree $d$ is a multiple of $3$, say $d=3e$. I sketched the argument in the comments, but now that comment has disappeared. For a point $x$ in $X$, there is a Grassmannian $G= \mathbb{G}(\mathbb{P}^3,\mathbb{P}^{n-1})$ parameterizing linear spaces $L\_x$ of dimension $4$ that contain $x$. If $x$ is a general point, then there is a dense open subset $U$ of $G$ parameterizing $L\_x$ that are not contained in $X$. There is a morphism,
$$
\zeta: U \to [\mathbb{P}k[x\_0,x\_1,x\_2,x\_3,x\_4]\_d / \textbf{PGL}\_5],
$$
that sends each $L\_x$ to the moduli point of the degree d hypersurface $L\_x\cap X$ in $L\_x \cong \mathbb{P}^4$. There is a subvariety $S$ of the target parameterizing hypersurfaces $[Y]$ that are of the form $e[Z]$ for a cubic hypersurface $Z$ (using additive notation). If the morphism $\zeta$ is surjective, then the image will intersect $S$, even if $e$ does not equal $1$.


It is a straightforward parameter count to determine when $\zeta$ should be dominant. That parameter count can be made into a rigorous proof. An example of this is in my preprint, "Fano Varieties and Linear Sections of Hypersurfaces". 


**Edit.** I was asked to provide more details. Let $F(x\_0,x\_1,x\_2,x\_3,x\_4)$ be any nonzero homogeneous polynomial of degree $d=3e$, e.g., $(x\_0^3+x\_1^3+x\_2^3+x\_4^3)^e$. Consider $F$ as an element in both $k[x\_0,\dots,x\_3]\_d$ and in $k[x\_0,\dots,x\_n]\_d$, the finite dimensional $k$-vector space of homogeneous polynomials $G$ of degree $d$ on $\mathbb{P}^4$, resp. on $\mathbb{P}^n$. Let $V\_F \subset k[x\_0,\dots,x\_n]\_d$ denote the subspace of those $G$ such that $G(x\_0,x\_1,x\_2,x\_3,0,\dots,0)$ is a scalar multiple of $F$. Note that $V\_F$ contains a subspace, call it $W$, of polynomials $G$ such that $G(x\_0,x\_1,x\_2,x\_3,0,\dots,0)$ is the zero polynomial. Moreover $W$ does not equal all of $V\_F$, since $F$ is in $V\_F\setminus W$. Thus $W$ is a codimension $1$ linear subspace of $V\_F$. Thus also the projective linear subspace $\mathbb{P}W$ of $\mathbb{P}V\_F$ has codimension $1$.


Now let $A$ denote the linear algebraic group of all projective linear automorphisms $\alpha$ of $\mathbb{P}^n$ that map the point $p=[1,0,\dots,0]$ to itself. Let $U\subset k[x\_0,\dots,x\_n]\_d$ denote the subspace of all $G$ such that $G(1,0,\dots,0)$ equals $0$.  

Since $\mathbb{P}W$ is a Cartier divisor in $\mathbb{P}V\_F$, also $A\times \mathbb{P}W$ is a Cartier division in $A\times\mathbb{P}V\_F$. There is an induced morphism,
$$
f : A\times \mathbb{P} V\_F \to \mathbb{P}U, \ \ f(\alpha,[G]) = G\circ \alpha^{-1}.
$$
Denote by $f\_W$ the restriction of $f$ to $A\times \mathbb{P}W$.


**Hypothesis.** The positive integers $(n,d)$ satisfy the inequality $4(n-4) \geq (d+3)(d+2)(d+1)/6$.


Under this hypothesis, the claim is that $f$ is surjective. To prove this, it suffices to prove that the restriction $f\_W$ is surjective. This new claim is equivalent to the claim that *every* point $q$ on every degree $d$ hypersurface $\text{Zero}(G)$ in $\mathbb{P}^n$ is contained in a linear $\mathbb{P}^4$ that is contained in $\text{Zero}(G)$. Assuming this claim, then we are done: let $S \subset \mathbb{P}U$ be the open subset parameterizing smooth hypersurfaces. Then $f^{-1}(S)$ is an open subset of $A \times \mathbb{P} V\_F$. It is nonempty by the claim. This nonempty open subset cannot be contained in the Cartier divisor $A\times \mathbb{P}W$. Thus, this nonempty open subset intersect $A\times \mathbb{P}V\_F \setminus A \times \mathbb{P}W$. That means that for every sufficiently general smooth hypersurface and for every general point on that hypersurface, there is a is a linear $\mathbb{P}^4$ containing that point whose intersection with the hypersurface is projectively linearly equivalent to $\text{Zero}(F)$.


So now it suffices to prove that $f\_W$ is surjective. This can be a bit delicate if we want to get precisely the correct inequality above. However, if you allow me to make $n$ much larger, then it is trivial. We can make the normal bundle of a given $\mathbb{P}^4$ in a given hypersurface "sufficiently general", via appropriate choices of the partial derivatives $\partial G/\partial x\_{4+i}|\_{\mathbb{P}^4}$, so that even the derivative of $f\_W$ is explicitly surjective.


