If 3x - 1 = 0 \(\Leftrightarrow\) x = \(\dfrac{1}{3}\) \(\Rightarrow\) The equality is always correct

If 3x - 1 \(\ne\) 0 \(\Rightarrow\) x² + 2 = 7x - 10

\(\Rightarrow\) x² - 7x + 2 = -10

\(\Leftrightarrow\) x² - 2 \(\times\) 3,5 \(\times\) x + (3,5)² - (3,5)² + 2 = -10

\(\Leftrightarrow\) (x - 3,5)² - 10,25 = -10

\(\Leftrightarrow\) (x - 3,5)² = 0,25 = (0,5)²

\(\Rightarrow\) \(\left[{}\begin{matrix}x-3,5=0,5\\x-3,5=-0,5\end{matrix}\right.\)\(\Rightarrow\) \(\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\) (3x - 1 \(\ne\) 0)

So x \(\in\) {\(\dfrac{1}{3}\);3;4}

\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)

\(\Leftrightarrow x^2+2=7x-10\)

\(\Leftrightarrow x^2+2-7x+10=0\)

\(\Leftrightarrow x^2-7x+12=0\)

\(\Leftrightarrow x^2-2.x.\dfrac{7}{2}+\left(\dfrac{7}{2}\right)^2-\dfrac{1}{4}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{2}\right)^2=\dfrac{1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{7}{2}=\dfrac{1}{2}\\x-\dfrac{7}{2}=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)

So  \(\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)