\( 1 \leftarrow \quad \) Solve the equation. \( (x-3)^{2 / 3}+(x-3)^{1 / 3}-6=0 \) The solution set is \( \{ \)
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To solve the equation \( (x-3)^{2/3} + (x-3)^{1/3} - 6 = 0 \), we can make a substitution to simplify it. Let \( y = (x-3)^{1/3} \). Then the equation rewrites to: \[ y^2 + y - 6 = 0. \] Now, we can factor the quadratic: \[ (y - 2)(y + 3) = 0. \] This gives us two solutions for \( y \): 1. \( y - 2 = 0 \) → \( y = 2 \) 2. \( y + 3 = 0 \) → \( y = -3 \) Now we can substitute back for \( x \): 1. For \( y = 2 \): \[ (x-3)^{1/3} = 2 \] Raising both sides to the power of 3 gives us: \[ x-3 = 2^3 = 8 \] Thus, \( x = 11 \). 2. For \( y = -3 \): \[ (x-3)^{1/3} = -3 \] Raising both sides to the power of 3 gives us: \[ x-3 = (-3)^3 = -27 \] Thus, \( x = -24 \). The solution set is \( \{ 11, -24 \} \).