College Algebra 7th Edition

$\dfrac{t^2+t+1}{t+1}$
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: $\require{cancel} \\=\dfrac{\cancel{(t-1)}(t^2+t+1)}{\cancel{(t-1)}(t+1)} \\=\dfrac{t^2+t+1}{t+1}$