1. The problem is to solve the equation $3\sqrt[n]{x}w + 1 = 1$ for $x$. 2. Start by subtracting 1 from both sides to isolate the term with $x$: $$3\sqrt[n]{x}w + 1 - 1 = 1 - 1 \Rightarrow 3\sqrt[n]{x}w = 0$$ 3. Since $w$ is multiplied by $3\sqrt[n]{x}$ and the product is zero, either $w=0$ or $\sqrt[n]{x} = 0$. 4. If $w \neq 0$, then $\sqrt[n]{x} = 0$. Recall that $\sqrt[n]{x} = x^{1/n}$, so this means: $$x^{1/n} = 0$$ 5. Raising both sides to the $n^{th}$ power to solve for $x$: $$\left(x^{1/n}\right)^n = 0^n \Rightarrow x = 0$$ 6. So, the solution is $x = 0$ provided $w \neq 0$. If $w=0$, then any $x$ satisfies the equation because it reduces to $1=1$. 7. Final answer: $$x = 0 \text{ if } w \neq 0 \text{ or any } x \text{ if } w = 0.$$