This is not an answer, but a guide to a simplification. The first congruence is fine. Note that the second congruence is equivalent to $x\equiv 5\pmod{8}$. Any solution of this congruence must be odd.
Now look at the third congruence. Note that as long as we know that $x$ is odd, we automatically have $5x\equiv 1\pmod{2}$. So in the presence of the second congruence, the third one can be replaced by $5x\equiv 1\pmod{9}$.
Thus we are looking at the congruences $3x\equiv 1 \pmod{7}$, $x\equiv 5\pmod{8}$, and $5x\equiv 1\pmod{9}$. Now the moduli are pairwise relatively prime. Relatively prime moduli are easier to handle, there is less risk of error.
There will be a unique solution modulo $(7)(8)(9)$. In particular, your final modulus of $126$ cannot be right.
It is probably worthwhile to simplify the congruences still further. Note that the congruence $3x\equiv 1\pmod{7}$ has the solution $x\equiv 5\pmod{7}$. And the congruence $5x\equiv 1\pmod{9}$ is equivalent to $x\equiv 2\pmod{9}$.
We got lucky, ended up with $x\equiv 5\pmod{7}$ and $x\equiv 5\pmod{8}$, which has as only solution $x\equiv 5\pmod{56}$.
So we are trying to solve $x\equiv 5\pmod{56}$, $x\equiv 2\pmod{9}$.
One can find a solution by a short search. Or else we want $x=56a+5=9b+2$. That gives $9b=56a+3$, so $3$ must divide $a$, say $a=3c$. We arrive at $56c+1=3b$. Clearly $c=1$ works, so $a=168$. The solution is therefore $x\equiv 173\pmod{(56)(9)}$.
Yes your method is correct according to the question .
Regarding the second part of your question,
You fetched
$n'\equiv0(mod9)$
$n'\equiv1(mod9)$
$n'\equiv8(mod9)$
$\forall n'\in\mathbb Z $and n' is the sum of 3 cubes
Now you see
The actual definition of congruences is
If $ m\equiv n(mod a)$
$Then ,$
$m=a.q_1+r$
And $b=a.q_2+r$
Where $ a,q_1,q_2,r\in \mathbb Z$
Or in other words two numbers are said to be congruent modulo a iff they leave same remaiders when divided by a
Now you see we say
If $m\equiv n(mod a)$,
Then $m\equiv n\pm k.a(mod a)$
PROOF:
Given:$m\equiv n(mod a)$
To show:$m\equiv n\pm k.a(mod a)$
proof:
If $m\equiv n(mod a)$.
Then m and n leaves same remainder with a.
Now,
$n=a.q_2+r$.
$\Rightarrow n\pm k.a=a.q_2+r\pm k.a$
$\Rightarrow n\pm k.a=a(q_2\pm k.a)+r$ where $k\in \mathbb Z$
Now ,
Since $q_2,k\in \mathbb Z$
And $q_2 \pm k.a\in \mathbb Z$
So we see now $n \pm a $ leaves same remainder as m and n
So we can say,
$m\equiv n\pm k.a (mod a)$
-×-×-×-×-×-×-×-×-×-×-
Using this theorem as we prove what you have asked
You found
$n'\equiv 8(mod9)$
Therefore $n'\equiv 8-9(mod 9)$
So we get $n'\equiv -1(mod 9)$
So writing 8(mod 9) and -1(mod 9) is all the same.
So you are done!!
Best Answer
You can write the equation as $(x +41)^{2} \equiv -27$ (mod 81). But this means there is no solution, as $x+41$ would need to be divisible by $9.$ I'll put in more explanation, in view of comments and questions, including correcting a sign error pointed out in the comments. We have $x^{2} + x + 7 \equiv x^{2}+ 82x + 7 $ (mod 81). This is (mod 81), equal to $(x+41)^{2}+ 7 -1600,$ and also $1600 \equiv -20$ (mod 81). Hence we need $(x+41)^{2} \equiv -27$ (mod 81), which is impossible in integers.