#### Answer

$c=-4,-1,1$

#### Work Step by Step

Given: $f(c)=c^3+4c^2-c+6$
Plug $f(c)=10$ then we have
$c^3+4c^2-c+6=10$
This can be written as: $c^3+4c^2-c-4=0$
Need to use factorization to solve the quadratic equation .
or, $(c+4)(c^2-1)=0$
$(c+4)(c+1)(c-1)=0$
This implies $(c+4)=0$
Now, $c=-4 $ and $c=-1,1$
Hence, $c=-4,-1,1$