$$y^2=x^3+Ax^2+Bx+C.$$ Change $x\longrightarrow x-A/3$, so that the new equation has the form $$y^2=x^3+ax+b.$$ Can you show step by step what operations we do in the first equation while obtaining the last equation?
Why did we use the $x\longrightarrow x-A/3$ exchange ?
the sum of the three roots of $p(x)=x^3+Ax^2+Bx+C$ is $-A.$
If $r_1,r_2,r_3$ are the roots of $p(x)$ then $r_1+\frac A3,r_2+\frac A3,r_3+\frac A3$ are the roots of $q(x)=p\left(x-\frac A3\right),$ so the sum of the roots of $q(x)$ is $0.$
Since $q$ is also a monic cubic polynomial (why?,) this means that $q(x)=x^3+ax+b$ for some $a,b.$