#### Answer

$(5m^3-7)(2m^3-3)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
10m^6-29m^3+21
,$ find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ Use these $2$ numbers to decompose the middle term of the given quadratic expression and then use factoring by grouping.
$\bf{\text{Solution Details:}}$
Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $
10(21)=210
$ and the value of $b$ is $
-29
.$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{
-14,-15
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
10m^6-14m^3-15m^3+21
.\end{array}
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
(10m^6-14m^3)-(15m^3-21)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
2m^3(5m^3-7)-3(5m^3-7)
.\end{array}
Factoring the $GCF=
(5m^3-7)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(5m^3-7)(2m^3-3)
.\end{array}