Question

How do you solve `root3( { 3x - 13}) = root3(x-21)`?

Answer 1

`x = -4`

Explanation:

`root(3)(3x-13) = root(3)(x-21)`

First, `color(blue)("cube")` both sides:

`(root(3)(3x-13))^color(blue)(3) = (root(3)(x-21))^color(blue)(3)`

The cube root and cube cancel out, so it becomes:
`3x - 13 = x - 21`

Now subtract `color(blue)x` from both sides;
`3x - 13 quadcolor(blue)(-quadx) = x - 21 quadcolor(blue)(-quadx)`

`2x - 13 = -21`

Add `color(blue)13` to both sides:
`2x - 13 quadcolor(blue)(+quad13) = -21 quadcolor(blue)(+quad13)`

`2x = -8`

Divide both sides by `color(blue)2`:
`(2x)/color(blue)2 = -8/color(blue)2`

Therefore,
`x = -4`

Hope this helps!

Answer 2

`color(maroon)(x = -4`

Explanation:

`"Given : " (3x-13)^(1/3) = (x - 21)^(1/3)`

`((3x-13)^(1/3))^3 = ((x-21)^(1/3))^3`

`(3x - 13)^cancel(color(red)(((1*3)/3))) = (x - 21)^cancel(color(red)(((1*3)/3)`

`3x - 13 = x - 21`

`3x - x = 13 - 21`

`2x = -8`

`color(maroon)(x = -4`