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The solutions to the equation

x^{2} u - x^{2} - u x + x + 6 - 6 u = 0

are $u = α, x = β$ or $x = γ$ , where $α, β, γ$ are all distinct.

Find the value of

(β γ)^{5} - α

Hint

Focus on factorising the $u$ out of three of the terms first. Then you can get an expression of the form

(u - 1) (\dots)

which should yield to simple factorisation.

Solution

Factorising out $u$ first, we get

\begin{aligned} x^{2} u - x^{2} - u x + x + 6 - 6 u \\ = & u (x^{2} - x - 6) - (x^{2} - x - 6) \\ = & (u - 1) (x^{2} - x - 6) \\ = & (u - 1) (x - 3) (x + 2) \end{aligned}