Cubed exponent equation 
$$\left(2 · 3^x\right)^3 + \left(9^x − 3\right)^3 = \left(9^x + 2 · 3^x − 3\right)^3$$ Solve the equation

I got the answer to the problem, in which I evaluated the whole expression (which was quite hard) which was $0$ and $1/2$, is there a way to solve this problem in a less tedious way?
 A: First of all, take $z=3^x$. That way, you obtain the equation:
$$(2z)^3 + (z^2 − 3)^3 = (z^2 + 2z − 3)^3.$$
Then try to factor it using your favourite method (Rufini, for instance), so that you obtain:
$$z (z - 1) (z + 3) (z^2 - 3)=0.$$
Solve for $z$ and use those to compute $x$ knowing that $z=3^x$.
A: Hint. The left side is the sum of two cubes, and
$$
a^3 + b^3 = (a+b)(a^2 -ab + b^2) .
$$
A: Using $t:=3^x$ and following the factorization by @Arthur ($(a+b)^3-a^3-b^3=3ab(a+b)$), the equation is equivalent to
$$t(t^2-3)(t^2+2t-3)=t(t-\sqrt 3)(t+\sqrt 3)(t-1)(t+3)=0.$$
Only the positive $t$ yield a solution and we immediately have
$$x=\frac12\text{ or }x=0.$$
A: Note that the entire equation is of the form
$$
a^3 + b^3 = (a+b)^3
$$
with $a = 2\cdot 3^x$ and $b = 9^x - 3$. This reduces to
$$
0 = 3a^2b + 3ab^2\\
0 = 3ab(a+b)
$$
which has the three solutions $a = 0$, $b = 0$, or $a+b = 0$. Clearly, $a = 0$ can't happen. $b = 0$ means $9^x = 3$, which has the solution $x = \frac12$. Finally, $a+b = 0$ means
$$
2\cdot 3^x + 9^x - 3 = 0\\
(3^x)^2 + 2\cdot 3^x - 3 = 0\\
3^x = \frac{-2\pm\sqrt{2^2 - 4\cdot 1\cdot (-3)}}{2} = -1\pm 2
$$
Since $3^x$ can't be negative, we can only keep $3^x = 1$, which gives $x = 0$.
A: HINT: You can use the identity
$$
x^3+y^3=(x+y)(x^2-xy+y^2)
$$
on the left hand side.
Preferably after you let $3^x=a$.
A: Some hints:
Let $u=3^x$. Then the equation becomes $(2u)^3+(u^2-3)^3=(u^2+2u-3)^3$. 
You can rearrange this and it factorises as $6u(u-1)(u+3)(u^2-3)=0$.
