Find two different ways to show that g(x)=-x^3+27 has only one x-intercept

An x intercept is where you have a point on the graph where y = 0, in othet words if you look at the graph, any points on the horizontal line going across the center: (x-axis), One way to show x intercepts is by actually showing the graph and circling where these points are, another, is to simplify the equation and find all of the x's when y is equal to zero. In this problem, there should only be 1. First I would replace g(x) with zero, so its 0 = -x^3+27 and subtract 27 on both sides: -27=-x^3, then divide by negative one on both sides to get rid of the negative in front of the x: 27=x^3, and now just cube root both sides, x^3 cubed is x and 27 cubed is 3, so your only x-intercept is 3

Sin(a)+cos(a)+cot(a)-1=2(csc(x)+1)