In the past or future

• We are given this expression:

$\dfrac{3^{2019} + 3^{2020} + 3^{2021}}{3^{2020} + 3^{2020} + 3^{2020}}$

• We can factor out $3^{2019}$

$\dfrac{\blue{3^{2019}}(1 + 3 + 3^2)}{\blue{3^{2019}}(3 + 3 + 3)}$

• $3^{2019}$ can be cancelled from the numerator and denominator

$\dfrac{\cancel{\blue{3^{2019}}}(1 + 3 + 3^2)}{\blue{\cancel{3^{2019}}}(3 + 3 + 3)} \implies \dfrac{1 + 3 + 9}{3 + 3 + 3}$

• Simplifying gives:

$\boxed{\dfrac{13}{9}}$

$\frac {3^{2019}+3^{2020}+3^{2021}}{3^{2020}+3^{2020}+3^{2020}} = \frac {3^{2019}(1+3+9)}{3^{2019}(3+3+3)} = \boxed{\frac {13}9}$

$\boxed{\frac{3^{2019} + 3^{2019} \times 3 + 3^{2019} \times 9}{3^{2019} \times 3 + 3^{2019} \times 3 + 3^{2019} \times 3}}$

$\boxed{\frac{3^{2019} \times (1+3+9)}{3^{2019} \times (3+3+3)}}$

$\boxed{\frac{13}{9}}$