How do I solve for X in the equation X^(1/3) - 4(X^[-1/3]) = 3?

[Doubell]

Homework Statement

SOLVE FOR X IN THE EQUATION:
X^(1/3) - 4(X^[-1/3]) = 3

Homework Equations

The Attempt at a Solution

I LET X^1/3 = Y
THEN THE ORIGINAL EQUATION BECOMES
Y - 4* 1/Y = 3
MULTIPLYING THE ENTIRE EQUATION BY Y RESULTS IN
Y^2 - 4 = 3Y
THEREFORE Y^2 -3Y - 4 = 0
THE ROOTS OF THE EQUATION ARE DEFINED BY (Y+1) (Y-4)
THEREFORE Y = -1 OR Y = 4
AND SINCE Y = X^1/3 THEN
X^1/3 = -1 OR X^1/3 = 4
THEREFORE [X^1/3]^3 = [-1]^3 SO X = -1 AND [X^1/3]^3 = [4]^3 SO X = 64
HENCE THE SOLUTIONS OF X ARE -1 AND 64.

 

[eumyang]

Homework Statement

SOLVE FOR X IN THE EQUATION:
X^(1/3) - 4(X^[-1/3]) = 3

STOP SHOUTING! (Please stop typing in all caps.)

The Attempt at a Solution

I LET X^1/3 = Y
THEN THE ORIGINAL EQUATION BECOMES
Y - 4* 1/Y = 3
MULTIPLYING THE ENTIRE EQUATION BY Y RESULTS IN
Y^2 - 4 = 3Y
THEREFORE Y^2 -3Y - 4 = 0
THE ROOTS OF THE EQUATION ARE DEFINED BY (Y+1) (Y-4)
THEREFORE Y = -1 OR Y = 4
AND SINCE Y = X^1/3 THEN
X^1/3 = -1 OR X^1/3 = 4
THEREFORE [X^1/3]^3 = [-1]^3 SO X = -1 AND [X^1/3]^3 = [4]^3 SO X = 64
HENCE THE SOLUTIONS OF X ARE -1 AND 64.

You're not done yet. Now you have to plug each solution into the original equation and check. By multiplying both sides by Y earlier you may have introduced an extraneous solution.

 

[Doubell]

ok for x= -1 ; (-1)^1/3 -4/(-1)^1/3 = 3 THIS IMPLIES -1 (-4)/-1 = -1 +4 = 3
AND FOR X = 64 ; (64^1/3) - 4/(64^1/3) = 3
WHICH IS 4 - 4/4 = 4-1 = 3 THE SOLUTIONS ARE TRUE FOR BOTH VALUES OF X. I WOULD LIKE TO SEE A DIFFERENT APPROACH THOUGH.

 

[SammyS]

STOP SHOUTING! (Please stop typing in all caps.)

I don't think he heard you !

 

[dynamicsolo]

ok for x= -1 ; (-1)^1/3 -4/(-1)^1/3 = 3 THIS IMPLIES -1 (-4)/-1 = -1 +4 = 3
AND FOR X = 64 ; (64^1/3) - 4/(64^1/3) = 3
WHICH IS 4 - 4/4 = 4-1 = 3 THE SOLUTIONS ARE TRUE FOR BOTH VALUES OF X. I WOULD LIKE TO SEE A DIFFERENT APPROACH THOUGH.

This really is the cleanest approach to solving this equation. You don't really want to "cube" both sides of the original equation, as this will just create more terms with various powers of the cube-root of x to be wrestled with.

While eumyang's suggestion to check for extraneous solutions is always in order, the danger doesn't really arise here. Since one can always take cube-roots of real numbers (and cube-root is "one-to-one"), there will not be a value of Y which does not correspond to a real value of x . Multiplying the transformed equation through by Y is also not an issue here, since we can see easily enough that Y = x^1/3 = 0 is not a solution to the equation, so we are not falsely introducing a solution or making the equation into potential nonsense.

 

[Doubell]

i apologize for the shouting.