Solve for $x$ in the equation: $\frac{x-a}{bc} + \frac{x-b}{ac} + \frac{x-c}{ab} = 2(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$

1. Problem Description

Solve for $x$ in the equation:

$\frac{x-a}{bc} + \frac{x-b}{ac} + \frac{x-c}{ab} = 2(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$

2. Solution Steps

Multiply both sides of the equation by $abc$:

$abc (\frac{x-a}{bc} + \frac{x-b}{ac} + \frac{x-c}{ab}) = abc (2(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}))$

Distribute $abc$ to each term on both sides:

$a(x-a) + b(x-b) + c(x-c) = 2bc + 2ac + 2ab$

Expand the terms:

$ax - a^2 + bx - b^2 + cx - c^2 = 2bc + 2ac + 2ab$

Combine terms with $x$:

$x(a+b+c) - (a^2 + b^2 + c^2) = 2bc + 2ac + 2ab$

Isolate the term with $x$:

$x(a+b+c) = a^2 + b^2 + c^2 + 2bc + 2ac + 2ab$

Recognize the right side as a perfect square:

$x(a+b+c) = (a+b+c)^2$

Divide both sides by $(a+b+c)$, assuming $a+b+c \neq 0$:

$x = \frac{(a+b+c)^2}{a+b+c}$

$x = a+b+c$

3. Final Answer

$x = a+b+c$