Answer
$\dfrac{t^2+t+1}{t+1}$
Work Step by Step
RECALL:
(1) $a^3-b^3=(a-b)(a^2+ab+b^2)$ (difference of two cubes)
(2) $a^2-b^2=(a-b)(a+b)$ (difference of two squares)
Factor the numerator using the formula for a difference of two cubes with $a=t$ and $b=1$, and factor the denominator using the formula for a difference of two squares with $a=t$ and $b=1$ to obtain:
$=\dfrac{(t-1)(t^2+t+1)}{(t-1)(t+1)}$
Cancel the common factors to obtain:
$\require{cancel}
\\=\dfrac{\cancel{(t-1)}(t^2+t+1)}{\cancel{(t-1)}(t+1)}
\\=\dfrac{t^2+t+1}{t+1}$
