Help writing proof for $\sqrt{a^2 + b^2} \neq \sqrt[3]{a^3 + b^3}$ Like the title says I'm attempting to write a existence proof showing "that there exists no non-zero real numbers a and b such that 
$\sqrt{a^2 + b^2} = \sqrt[3]{a^3 + b^3}$". 
I'm having trouble finding a starting point. I've tried manipulating the equation but haven't managed to isolate a or b. But I don't know if this helps me or where to go from there. 
If anybody could give a hint I would very much appreciate it. 
P.S. I would ask that you refrain from posting a full solution as I'm trying to learn this material. 
 A: Hint: Take the sixth power of both sides.
Added: (for completeness, after the OP had used the hint to solve the problem) Since $\sqrt{a^2+b^2}$ is by definition non-negative, we have
$$\sqrt{a^2+b^2}=\sqrt[3]{a^3+b^3} \qquad\text{iff}\qquad \left(\sqrt{a^2+b^2}\right)^6=\left(\sqrt[3]{a^3+b^3}\right)^6.$$
Equivalently, we want to find the solutions of
$$(a^2+b^2)^3=(a^3+b^3)^2.$$
in non-zero reals. Expand. We get 
$$a^6 +3a^4b^2+3a^2b^4+b^6=a^6+2a^3b^3+b^6,$$
which is equivalent to
$$3a^4b^2+3a^2b^4=2a^3b^3.$$
Since we are looking for solutions where neither $a$ nor $b$ is equal to $0$,  we are looking for non-zero real solutions of
$$3a^2+3b^2=2ab.$$
This equation has no non-zero real solutions. For by completing the square we get 
$$a^2+b^2-\frac{2ab}{3}=\left(a-\frac{b}{3}\right)^2+\frac{8b^2}{9}.$$
In order for the right hand side to be $0$, both terms must be $0$. In particular, $b$ must be $0$, and therefore so must $a$. 
Comment: You may find the following related idea interesting. Let $p>1$, and let $t$ be positive. We will prove  that 
$$1+t> (1+t^p)^{1/p}.  \qquad\qquad(\ast)$$
 Let $f(t)=1+t -(1+t^p)^{1/p}$. Note that $f(0)=0$.  So it is enough to show that for positive $t$, $f(t)$ is an increasing function.  To do this, we use the derivative: 
$$f'(t)=1 -\frac{t^{p-1}}{(1+t^p)^{(p-1)/p}}.$$
For positive $t$, $f'(t)$ is positive, since $(1+t^p)^{(p-1)/p}>(t^p)^{(p-1)/p}=t^{p-1}$. This completes the proof of $(\ast)$. 
To apply $(\ast)$ to our problem, observe first that a solution of our equation with positive $a$ yields a solution of  $(1+(b/a)^2)^{1/2}=(1+(b/a)^3)^{1/3}$.  It is easy to check that $b/a$ cannot be negative. Now let $t=(b/a)^2$. We obtain the equation $1+t=(1+t^{3/2})^{2/3}$, which, by  $(\ast)$, cannot hold for positive $t$.  
We reduced the problem to one variable in order to use familiar tools. But there are important generalizations to several variables.
A: If $a,b \in \mathbb R \setminus \{0\}$ then
$$\sqrt{a^2 + b^2} \neq \sqrt[3]{a^3 + b^3}\tag 1$$
Let $c = \dfrac ba$. Since $a \ne 0$, then $c$ is a real number and 
$$\sqrt{a^2 + b^2} = |a|\sqrt{1 + c^2}$$ and 
$$\sqrt[3]{a^3 + b^3} = a\sqrt[3]{1 + c^3}$$
So $(1)$ is equivalent to 
$$|a|\sqrt{1 + c^2} \neq a \sqrt[3]{1 + c^3}\tag{2}$$
We can prove $(2)$ by showing that
$|a|\sqrt{1 + c^2} = a \sqrt[3]{1 + c^3}$ leads to a contradiction. I will leave that up to you.
