Let $\xi$ be an $n$-dimensional complex vector bundle. It is claimed that the Chern class of $\xi$ is
$$
c(\xi)=(1+x_1)\cdots (1+x_n),
$$
$|x_k|=2$, $c_j(\xi)$ is the $j$-th symmetric polynomial of $x_1,\cdots,x_n$. Is this claim true?
In the case $\xi=L_1\oplus\cdots\oplus L_n$, Whitney sum of complex line bundles, I have obtained the claim. How about general case?
This is not true.
For example, consider $T\mathbb{CP}^2$ which has total Chern class $c(T\mathbb{CP}^2) = 1 + 3x + 3x^2$ where $x = c_1(\mathcal{O}(1)) \in H^2(\mathbb{CP}^2; \mathbb{Z})$ is a generator. Suppose $c(T\mathbb{CP}^2) = (1 + x_1)(1 + x_2)$ for some $x_1, x_2 \in H^2(\mathbb{CP}^2; \mathbb{Z})$, then $x_1 = kx$ and $x_2 = lx$ for some integers $k$ and $l$. Then $$1 + 3x + 3x^2 = (1 + kx)(1 + lx) = 1 + (k + l)x + klx^2.$$ This is impossible: if $k + l = 3$, one of $k$ and $l$ is even, but then $kl$ would be even, so it can't be equal to $3$.
This shows that $T\mathbb{CP}^2$ is not isomorphic to the direct sum of line bundles. Note however that if the total Chern class did factor, it does not necessarily mean the bundle is isomorphic to a direct sum; see this MathOverflow question.
What is true however is the splitting principle:
Note that
$$p^*c(\xi) = c(p^*\xi) = c(L_1\oplus\dots\oplus L_n) = c(L_1)\dots c(L_n) = (1 + x_1)\dots(1+x_n)$$
where $x_i = c_1(L_i)$.