Struggling with this question: $15x^3 -6x^2 + 2x +26 \cong 0 \mod343$.
Here is what I have so far: By Hensel's Lemma if we have a solution to $f(x) \cong 0\mod p$ we can find solution to $f(x) \cong 0\mod p^\alpha$
Note $343=7^3$.
So we must consider: $15x^3 -6x^2 + 2x +26 \cong 0 \mod7$
But I have no idea how to solve this latter equation.
On
Another answer.-Since $7$ is small we can simply make a table (not advisable of course for big $p$). Because, in $\mathbb F_7$, $$15x^3 -6x^2 + 2x +26=x^3+x^2+2x-2$$ we note, for short, $f(x)=x^3+x^2+2x-2$ so we have
$$(\mathbb F_7,f(x))\in\{(1,2),(2,0),(3,5),(4,2),(5,4),(6,3),(7,5)\}$$ where we see that the only solution is $x=2$ modulo $7$.
Consequently there is solution modulo $343$. Calculation is now hard: Wolfram gives $\boxed{x=240}$ as minimal integer positive solution and the only thing I do is to verify this solution.
Anyway we have, concisely written, $$f(x)\equiv0\pmod7\Rightarrow x=2\\f(2+7x)\equiv0\pmod{7^2}\Rightarrow x=6\\f(2+7\cdot6+7^2\cdot x)\equiv0\pmod{7^3}\Rightarrow x=4$$
Consequently the solution is equal to $$x=2+7\cdot6+7^2\cdot4=240$$
You can solve it by noting that $$ 15x^3-6x^2+2x+26=(x^2+3x+1)(x+5) $$ over $\Bbb F_7$. The quadratic polynomial is irreducible and has no root in $\Bbb F_7$. So the only solution is $x=-5=2$.