\sqrt[3]{ 3x + 1 - 4 = - 6}

Question

`\sqrt[3]{ 3x + 1 - 4 = - 6}`

Answer 1

i'm not sure if -6 is out or inside the cubic root but i will solve it for both cases

In case of
`\sqrt[3]{3x+1-4} =sqrt[3][-6]` --> (by multiplying the power by 3)

`3x+1-4 = -6`

`3x -3 = -6` ---> by adding 3 for the both sides

`3x = -3` ---> by dividing both sides by 3

`x = -1`

in case of
`\sqrt[3]{3x+1-4} = -6` ----> we will repeat the first step again
therefore,

`3x + 1 -4 = (-6)^(3) = -216`

`3x -3 = -216` ----> by adding both sides by 3

`3x = -213` ----> by dividing both sides by 3

`x = (-213)/(3) = -71`
then x = -71
I hope that helps ;)