How do you solve #x^ { 2/ 3} - x ^ { 1/ 3} = 12#?

Question

505 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-16T15:58:23

#x=64,-27#

#x^(2/3)-x^(1/3)=12--(1)#

let #u=x^(1/3)#

#(1)rarru^2-u=12#

a quadratic in #u#

#:.u^2-u-12=0#

#(u-4)(u+3)=0#

#u+3=0=>u=-3#

ie

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

#:.x=(-3)^3=-27#

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

#:.x=4^3=64#

two possible solutions

#x=64, -27#