first step: put 3^x in factor:
3^x*x^2-3^x*x = 3^x(x^2-x)
so: 3^x(x^2-x)=0 and you know that 0/x is always 0. so you divide both sides by 3^x, (and we just established 0/3^x=0) so you have :
x^2-x=0. then you add x to both sides,
x^2=x. now, there are two solutions possible. here you have to find the two values of x who stay the same when they are squared. these two values are x=0 and x=1.
so: 1^2=1 ; and 0^2=0.
