A function takes in an input and returns an output.

In this function #g(x)#, you input #x# and it returns #x^3-2x^2#. We can also look at this as this: you input #x# (what's enclosed by the parentheses) and this function #g(x)# takes the input (x), cubes it (#x^3#), subtracts two times the input squared (#-2x^2#). Putting this all together, this particular function #g(x)# takes in an input #x# and returns the expression #x^3-2x^2#, with #x# representing the input (which could be anything).

Cool! So what exactly does the question mean by #g(x+1)#? What this means is that the entire expression #x+1# is the input! So, because

#g(x)=x^3-2x^2#

so,

#g(x+1)=(x+1)^3-2(x+1)^2#

So now, all we have to do is use algebra to simplify this expression! We have

#g(x+1)=(x+1)^3-2(x+1)^2#

#g(x+1)=(x+1)(x+1)(x+1)-2(x^2+2x+1)#

#g(x+1)=(x+1)(x^2+2x+1)-2(x^2+2x+1)#

Here, we see that we have a #(x^2+2x+1)# term in common on both sides of the minus sign. We invoke the distributive property:

#g(x+1)=(x^2+2x+1)(x+1-2)#

#g(x+1)=(x^2+2x+1)(x-1)#

And we're done!

SIDENOTE:

Alternatively, you could also write the answer like this:

#g(x+1)=(x+1)^2(x-1)=(x^2-1)(x+1)#

Either way is fine.