How do you multiply $(t - 3) (t^{2} + 3 t + 9)$ $(t-3)(t^2+3t+9)$ ?

Question

1756 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-27T06:29:34

$= t^{3} - 27$ $=t^3-27$

$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ $a^3-b^3=(a-b)(a^2+ab+b^2)$

Similarly

$(t - 3) (t^{2} + 3 t + 9)$ $(t-3)(t^2+3t+9)$ can be written as

$t^{3} - 3^{3}$ $t^3-3^3$

$= t^{3} - 27$ $=t^3-27$

How do you multiply (t-3)(t^2+3t+9)(t−3)(t2+3t+9)(t-3)(t^2+3t+9)?