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Algebra Difficulty 5.0 AIME Prove it Bulgaria

Let aa be the largest value of the expression 24y−9y224y-9y^2, where yy is a rational number and bb is the smallest integer satisfying the inequality
(t+3)3−(6t−7)2−(t−9)3<3. (t+3)^3 - (6t-7)^2 - (t-9)^3 < 3.
Factor into irreducible factors with integer coefficients the expression
a(x−1)x3+bx−2x−1. a(x-1)x^3 + bx - 2x - 1.

Solution

We have 24y−9y2=16−(3y−4)224y - 9y^2 = 16 - (3y - 4)^2 whose largest value a=16a = 16 is reached for y=43y = \frac{4}{3}. The given inequality is equivalent to
t3+9t2+27t+27−36t2+84t−49−t3+27t2−243t+729<3−132t+704<0, \begin{aligned} & t^3 + 9t^2 + 27t + 27 - 36t^2 + 84t - 49 - t^3 + 27t^2 - 243t + 729 < 3 \\ & -132t + 704 < 0, \end{aligned}
i.e. t>163t > \frac{16}{3} and b=6b = 6. Substituting a=16a = 16, b=6b = 6 in the given expression, we get
16(x−1)x3+6x−2x−1=16x4−16x3+4x−1=(16x4−1)−4x(4x2−1)=(4x2−1)(4x2+1)−4x(4x2−1)=(4x2−4x+1)(2x−1)(2x+1)=(2x−1)3(2x+1). \begin{aligned} & 16(x-1)x^3 + 6x - 2x - 1 = 16x^4 - 16x^3 + 4x - 1 \\ & = (16x^4 - 1) - 4x(4x^2 - 1) = (4x^2 - 1)(4x^2 + 1) - 4x(4x^2 - 1) \\ & = (4x^2 - 4x + 1)(2x - 1)(2x + 1) = (2x - 1)^3(2x + 1). \end{aligned}

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