Answer
$t(0.9+t)(0.9-t)$
Work Step by Step
Factoring the $GCF=
t
$, the given expression, $
0.81t-t^3
,$ is equivalent to
\begin{array}{l}
t(0.81-t^2)
.\end{array}
Using $x^2-y^2=(x+y)(x-y)$ or the factoring of the difference of 2 squares, then the factored form of the expression, $
t(0.81-t^2),$ is
\begin{array}{l}
t(0.9+t)(0.9-t)
.\end{array}
