If $(a + b\sqrt c)^n = d + e\sqrt c$, then $(a - b\sqrt c)^n = d - e\sqrt c$ I think that:

If $a,b,(c\ge0$ not a prefect square$),d,e,f\in\mathbb Z$ such that for some $n \ge 1$, $(a + b\sqrt c)^n = d + e\sqrt c$, then $(a - b\sqrt c)^n = d - e\sqrt c$

Is this true? Can someone provide a proof or give a hint for how to prove this? (preferably without induction if it doesn't provide insight)

Since $\sqrt{c}$ is irrational, then
$$
(a+b\sqrt{c})^n=\sum_{i=0}^n \binom{n}{i}a^{n-i}b^ic^{i/2}=\underbrace{\sum_{i \text{even}}\binom{n}{i}a^{n-i}b^ic^{i/2}}_{d}+\sqrt{c}\underbrace{\sum_{i\text{ odd}}\binom{n}{i}a^{n-i}b^ic^{(i-1)/2}}_{e}.
$$
On the other hand
$$
(a-b\sqrt{c})^n=\sum_{i \text{even}}\binom{n}{i}a^{n-i}(-b)^ic^{i/2}+\sqrt{c}\sum_{i\text{ odd}}\binom{n}{i}a^{n-i}(-b)^ic^{(i-1)/2}=d-e\sqrt{c}.
$$
