a) \(\left(4x-3\right)^3+\left(3x-2\right)^3=\left(7x-5\right)^3\)
\(\Leftrightarrow64x^3-144x^2+108x-27+27x^3-54x^2+36x-8=343x^3-735x^2+525x-125\)
\(\Leftrightarrow-252x^3+537x^2-381x+90=0\)
\(\Leftrightarrow-3\left(84x^3-179x^2+127-30\right)=0\)
\(\Leftrightarrow-3\left(7x-5\right)\left(3x-2\right)\left(4x-3\right)=0\)
\(\Leftrightarrow x\in\left\{\frac{5}{7};\frac{2}{3};\frac{3}{4}\right\}\)
b) \(x^3-2x^2-x-6=0\)
\(\Leftrightarrow x^3-3x^2+x^2-3x+2x-6=0\)
\(\Leftrightarrow x^2\left(x-3\right)+x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+x+2\right)=0\)
Vì \(x^2+x+2>0\)
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy....
