\(\left(x-1\right)^3-\left(3x-5\right)\left(3x+5\right)=\left(x-3\right)\left(x^2+3x+9\right)-3x\left(x-1\right)-9x^2+x\)
\(\Leftrightarrow x^3-3x^2+3x-1-9x^2+25=x^3-27-3x^2+3x-9x^2+x\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(-3x^2-9x^2+9x^2+3x^2\right)+\left(3x-3x-x\right)-1+25+27=0\)
\(\Leftrightarrow-x+51=0\)
\(\Leftrightarrow x=51\)
Vậy ,....