# How do you solve #x^(2/3) - 3x^(1/3) - 4 = 0#?

##### 2 Answers

Set

Roots are

#### Explanation:

Set

So the equation becomes:

To solve for

x = 64 or x = -1

#### Explanation:

note that

# (x^(1/3))^2 = x^(2/3) # Factorising

# x^(2/3) - 3x^(1/3) - 4 = 0 # gives ;

# ( x^(1/3) - 4 )( x^(1//3) + 1 ) = 0#

#rArr ( x^(1/3) - 4 ) = 0 or ( x^(1/3) + 1 ) = 0#

#rArr x^(1/3) = 4 or x^(1/3) = - 1 #

'cubing' both sides of the pair of equations :

# (x^(1/3))^3 = 4^3 and (x^(1/3))^3 = (- 1 )^3 #

#rArr x = 64 or x = - 1 #