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1 solution
Surprisingly, it is constant! Let k = 3 4 − 3 x + 1 6 − 2 4 x + 9 x 2 − x 3 + 3 4 − 3 x − 1 6 − 2 4 x + 9 x 2 − x 3 Using ( a + b ) 3 = a 3 + b 3 + 3 ( a + b ) a b , k 3 = 8 − 6 x + 3 k 3 ( 4 − 3 x ) 2 − ( 1 6 − 2 4 x + 9 x 2 − x 3 ) ⇒ k 3 − 8 + 6 x − 3 k 3 x 3 = 0 ⇒ k 3 − 3 x k + 6 x − 8 = 0 ⇒ ( k − 2 ) ( k 2 + 2 k + 4 − 3 x ) = 0 ⇒ k = 2 or − 1 ± 3 x − 3 Note that − 1 ± 3 x − 3 are extraneous solutions.
Thus k = 2 for − 1 ≤ x ≤ 1 !