( x + 2 )3 - ( 2x + 3 )2 + ( 2x + 3 )( 2x - 3 ) = ( x - 2 )( x2 + 2x + 4 ) - 6x( x + 2 )
⇔ x3 + 6x2 + 12x + 8 - ( 4x2 + 12x + 9 ) + 4x2 - 9 = x3 - 8 - 6x2 - 12x
⇔ x3 + 10x2 + 12x - 1 - 4x2 - 12x - 9 = x3 - 6x2 - 12x - 8
⇔ x3 + 6x2 - 10 = x3 - 6x2 - 12x - 8
⇔ x3 + 6x2 - 10 - x3 + 6x2 + 12x + 8 = 0
⇔ 12x2 + 12x - 2 = 0
⇔ 2( 6x2 + 6x - 1 ) = 0
⇔ 6x2 + 6x - 1 = 0 (*)
Δ = b2 - 4ac = 62 - 4.6.(-1) = 60
Δ > 0 nên (*) có hai nghiệm phân biệt
\(\hept{\begin{cases}x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-6+\sqrt{60}}{12}=\frac{-3+\sqrt{15}}{6}\\x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-6-\sqrt{60}}{12}=\frac{-3-\sqrt{15}}{6}\end{cases}}\)
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