How do you solve #x^ { 2/ 3} - 6= x ^ { 1/ 3}#?

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Answer 1 · 2016-11-06T11:23:04

Solution : #x=27 or x=-8#

Let #x^(1/3)=z#

# :. x^(2/3)-6=x^(1/3)#

This can be also be written as:

#z^2-6=z" or " z^2-z-6=0#

which can be factored to give:

#(z-3)(z+2)=0#

# :. z-3 =0 or z+2=0#

# :. z=3 or z=-2#

Replacing value of #z =x^(1/3)# in x we get

#x^(1/3)=3 or x=27#

# and x^(1/3)=-2 or x= -8 #

Solution : #x=27 or x=-8# [Ans]