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\[ \frac{x - 3}{x + 3} - \frac{x + 3}{x - 3} + 6\frac{6}{7} = 0 \] Or, \[ \frac{(x - 3)^2 - (x + 3)^2}{(x + 3)(x - 3)} = -6\frac{6}{7} \] Or, \[ \frac{(x^2 - 6x + 9) - (x^2 + 6x + 9)}{x^2 - 9} = -\frac{48}{7} \] Or, \[ \frac{x^2 - 6x + 9 - x^2 - 6x - 9}{x^2 - 9} = -\frac{48}{7} \] Or, \[ \frac{-12x}{x^2 - 9} = -\frac{48}{7} \] Or, \[ \frac{x}{x^2 - 9} = \frac{4}{7} \] Cross-multiplying: \[ 4x^2 - 36 = 7x \Rightarrow 4x^2 - 7x - 36 = 0 \] Now factorizing: \[ 4x^2 - (16 - 9)x - 36 = 0 \Rightarrow 4x^2 - 16x + 9x - 36 = 0 \Rightarrow 4x(x - 4) + 9(x - 4) = 0 \Rightarrow (x - 4)(4x + 9) = 0 \] ∴ Either \(x - 4 = 0\) ⟹ \(x = 4\) Or \(4x + 9 = 0\) ⟹ \(x = -\frac{9}{4}\) ∴ The required solutions are \(x = 4\) or \(x = -\frac{9}{4}\)(Answer)