$pt\Leftrightarrow \left[(4x^3-x+3)^3- \frac 34 \right]-(x^3+ \frac 34)=0$$\Leftrightarrow (4x^3-x+3- \sqrt[3]{\frac 34} ) \left[(4x^3-x+3)^2+ \sqrt[3]\frac 34 (4x-x+3)+(\sqrt[3]{\frac 34})^2\right]- \frac {4x^3+3}4=0$(*)
Đặt $(4x^3-x+3)^2+ \sqrt[3]\frac 34 (4x-x+3)+(\sqrt[3]{\frac 34})^2=A$
Dễ chứng minh $A \ge \frac 34.(\sqrt[3]{\frac 34})^2> \frac 12$
$(*)\Leftrightarrow \left[ (4x^3+3)-(x + \sqrt[3]{ \frac 34})\right].A-\frac {4x^3+3}4=0$
$\Leftrightarrow \left[ (4x^3+3)- \frac{x^3+\frac 34}{B} \right].A -\frac {4x^3+3}4=0$
$\Leftrightarrow (4x^3+3)(A-\frac{A}{4B}- \frac 14)=0$
Với $B=x^2-\sqrt[3]{ \frac 34}.x+ (\sqrt[3]{\frac 34})^2 \ge \frac 34.(\sqrt[3]{\frac 34})^2\Rightarrow 4B>2$
Ta chứng minh $A - \frac{A}{4B} - \frac 14 > 0$
$\Leftrightarrow A. \frac{4B-1}{4B}- \frac 14>0$
Do $4B >2\Rightarrow \frac{4B-1}{4B}> \frac 12$
Và $A> \frac 12$
Nên pt có nghiệm duy nhất $x = - \sqrt[3]{\frac 34}$