i'm not sure if -6 is out or inside the cubic root but i will solve it for both cases
In case of [tex] \sqrt[3]{3x+1-4} =sqrt[3][-6][/tex] --> (by multiplying the power by 3)
[tex] 3x+1-4 = -6[/tex]
[tex] 3x -3 = -6 [/tex] ---> by adding 3 for the both sides
[tex] 3x = -3 [/tex] ---> by dividing both sides by 3
[tex] x = -1 [/tex]
in case of [tex] \sqrt[3]{3x+1-4} = -6[/tex] ----> we will repeat the first step again
therefore,
[tex] 3x + 1 -4 = (-6)^{3} = -216 [/tex]
[tex] 3x -3 = -216 [/tex] ----> by adding both sides by 3
[tex] 3x = -213 [/tex] ----> by dividing both sides by 3
[tex] x = \frac{-213}{3} = -71 [/tex]
then x = -71
I hope that helps ;)
