What is my answer?

Set $x^{3}-21x^{2}+148x-343$ equal to $x$ because they are both the answer to this problem and are therefore equal. Subtracting $x$ from both sides yields $x^{3}-21x^{2}+147x-343=0$ , which is the same as $(x-7)^{3}=0$ . Thus, the answer is $7$ .

$x^3 - 21x^2 + 148x - 343 = x \implies x^3 - 21x^2 + 147x - 343 = 0$ . Now assume $f(x) = x^3 - 21x^2 + 147x - 343 = 0$ .

$\begin{aligned} f(x) & = 0 \\ \Rightarrow x^3 - 21x^2 + 147x - 343 & = 0 \\ \Rightarrow x^3 - 7x^2 - 14x^2 + 98x + 49x - 343 & = 0 \\ \Rightarrow x^2(x-7)-14x(x-7)+49(x-7) & = 0 ~~~~~[\text{Since f(7) = 0, x - 7 is a factor of f(x)}] \\ \Rightarrow (x - 7)(x^2 - 14x + 49) & = 0 \\ \Rightarrow (x - 7)(x - 7)^2 & = 0 \\ \Rightarrow (x - 7)^3 & = 0 \\ \Rightarrow x - 7 & =0 \\ \implies x & = \boxed 7\\ \end{aligned}$

Relevant wiki: Rational Root Theorem - Basic

$\begin{aligned} x^3-21x^2+148x - 343 & = x & \small \color{#3D99F6} \text{As given} \\ x^3-21x^2+147x - 343 & = 0 & \small \color{#3D99F6} \text{By rational root theorem} \\ x^3-3(7)x^2+3(7^2)x - 7^3 & = 0 & \small \color{#3D99F6} \text{and noting that }343=7^3 \\ (x-7)^3 & = 0 \\ x & = \boxed 7 & \small \color{#3D99F6} \text{The only real solution} \end{aligned}$