Math problem about secant lines

The graph g(x)=x^3-2 passes through point T(1,-1). Determine an algebraic expression that represents the slope of any secant line TS that passes through T and S(x, x^3-2).

I tried this using the slope formula and my answer is (x^3-1)/(x-1), but the correct answer is suppossed to be 1+x+x^2. Can anyone explain how they got this?

Your answer is perfectly correct. Just do the division. Either synthetic division or long division will do it.

Long division :

. . . . . . . . . . x². + . x . + . .1
. . . . . . . . -------------------------------
. . . .x - 1.) . x³ .+. 0x². + . 0x . - . 1
. . . . . . . . . .x³ . .- . x²
. . . . . . . . . ------------
. . . . . . . . . . . . . . . x² . + . 0x
. . . . . . . . . . . . . . . x². . - . . x
. . . . . . . . . . . . . .. --------------
. . . . . . . . . . . . . .. . . . . . . . x. .- . 1
. . . . . . . . . . . . . .. . . . . . . . x . -. .1
. . . . . . . . . . . . . .. . . . . . . ------------
. . . . . . . . . . . . . .. . . . . . . . 0 . . . 0

So (x³ - 1) / (x - 1) = x² + x + 1

My answer was like yours, but a bit different.

I came up with (1 - x^3) / (1 - x).

I compared that formula with random values of x against what you said is the correct answer.

I get the same answer with both formulas. You formula is still right. Yours is actually easier since mine deals with negative numbers and yours keep them positive.

y - y1 = m(x - x1)

-1 - (x^3 - 2) = m(1 - x)
-1 - x^3 + 2 = m(1-x)
1 - x^3 = m(1-x)
m = (1-x^3) / (1-x)