How to graph and solve this equation? I'am trying to solve this equation.
\begin{equation}
x^{8}+(x+2)^{8}=2
\end{equation}
What I tried:
\begin{equation}g(x)=x^{8}+(x+2)^{8}\end{equation}
\begin{equation}
\begin{array}{l}
\text {  }\\
y=x+1
\end{array}
\end{equation}
\begin{equation}
(y+1)^{8}+(y-1)^{8}=2
\end{equation}
\begin{equation}
(y+1)^{8}=y^{8}+a_{7} y^{7}+a_{6} y^{6}+\cdots+a_{1} y+1
\end{equation}
\begin{equation}
(y-1)^{8}=y^{8}-a_{7} y^{7}+a_{6} y^{6}+\cdots-a_{1} y+1
\end{equation}
\begin{equation}
\begin{array}{l}
2\left(y^{8}+a_{6} y^{6}+a_{4} y^{4}+a_{2} y^{2}+1\right)=2 \Leftrightarrow \\
y^{8}+a_{6} y^{6}+a_{4} y^{4}+a_{2} y^{2}=0 \Leftrightarrow y^{2}=0 \Leftrightarrow y=0
\end{array}
\end{equation}
y=0, x= -1
Maybe there is another way of solving this...
 A: OK then, let's use calculus.
Consider the function $f(x) = x^8 + (2 + x)^8$. Note that $f(x) \to \infty$ as $x \to \pm \infty$, so the function must achieve a global minimum at a stationary point, i.e. where $f'(x) = 0$.
We have,
$$f'(x) = 8x^7 + 8(x + 2)^7.$$
This is $0$ if and only if
$$8x^7 + 8(x + 2)^7 = 0 \iff (x + 2)^7 = -x^7 \iff \left(\frac{x + 2}{x}\right)^7 = -1.$$
Note, for the above to work, we must observe that $x = 0$ is definitely not a solution. As we are considering only the reals, we can take the unique seventh root of both sides to obtain
$$\frac{x + 2}{x} = -1 \iff x + 2 = -x \iff x = -1.$$
That is, there exists one and only one stationary point for the whole function: at $x = -1$. Therefore, the global minimum of $f$ must be achieved only at this point. That is,
$$2 = f(-1) \le f(x)$$
for all $x \in \Bbb{R}$, with equality if and only if $x = -1$. This implies $f(x) = 2$ if and only if $x = -1$.
A: I encourage you to keep working on your solution using derivatives. However, solving this equation does not need derivatives. Here I write a hint and if you need more help I'll give you another one!
Hint1:

 Rearrange the equation: $(x+2)^8 - 1 = 1 - x^8$

Edit: Ah! Now that you posted your work I see you don't need help. That is a good solution. Well done!
Now, here is an alternative solution. Not as nice and concise as yours, but still might be a good exercise. Like you, I too felt more comfortable with the variable change $y = x + 1$ . In fact this would be the hint number two! We have:
$$(y+1)^8 - 1^8 = 1^8 - (y-1)^8$$
$$[(y+1)^4+1][(y+1)^2+1](y+2)y = [1+(y-1)^4][1+(y-1)^2]y(2-y)$$
Clearly, one answer is $y=0$, which means $x=-1$ .
Below, we show that the equation does not have any other real answers. Assuming that $y \ne 0$ we can divide both sides by $y$ :
$$[(y+1)^4+1][(y+1)^2+1](y+2) = [1+(y-1)^4][1+(y-1)^2](2-y)$$
Without any more factorisation, we can find several useful things in the above equation:
Firstly, we can see that if $y>2$ or $y<-2$ then the two sides of the equation have different signs. Therefore the equation does not have any solutions in those value ranges.
Secondly, at $y=2$ and at $y=-2$ one side of the equation becomes zero while the other side is nonzero. So these two are not solutions of the equation either.
Finally, for $-2 < y < 2$ where both sides are positive, we can see that if $y > 0$ then each term of the left hand side is greater than its corresponding term in the right hand side (therefore left hand side $ > $ right hand side); and if $y < 0$ then each term of the left hand side is less than its corresponding term in the right hand side (therefore left hand side $ < $ right hand side). Either way, the two sides cannot be equal.
So we are left with the only case that $y=0$.
A: Let's try
$$ 1 + 1 = 2.$$
$$ \left\{ 
\begin{aligned}
x^8=1\\
(x+2)^8=1 
\end{aligned}
\right.
$$
A solution is $x=-1.$

Actually, $x=-1$ is a double root:
Let $q(x) = x^8 + (x+2)^8 -2$. So we have $$q(x)=0.$$  We can factor q(x):
$$q(x) = 2 (x+1)^2 (127 + 258 x + 253 x^2 + 132 x^3 + 43 x^4 + 6 x^5 + x^6).$$
The equation $127 + 258 x + 253 x^2 + 132 x^3 + 43 x^4 + 6 x^5 + x^6=0$
has no real solutions.
A: $$f(x)=x^8+(x+2)^8\qquad f'(x)=8x^7+8(x+2)^7\qquad f''(x)=56x^6+56(x+2)^6$$
$f''(x)$is clearly positive so $f$ is convex and above its tangents, in particular since $f'(-1)=0$ it is above the horizontal line $y=f(-1)+(x-1)f'(-1)=2$.
Since this value is effectively reached for $x=-1$ then it is the global minimum (no other because of convexity), and you can generalize to any $2n$ exponent instead of $8$.
