How to solve #x^(2/3) - 6x^(1/3) + 5 = 0#? (substituting using u)

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Answer 1 · 2017-10-26T23:02:48

I got #x = 125# or #x = 1#.

We have

#(x^(1/3))^2 - 6x^(1/3) + 5 = 0#

Then letting #u = x^(1/3)#.

#u^2 - 6u + 5 = 0#

#(u - 5)(u - 1) = 0#

#u = 5 or 1#

Now we reverse the substitution.

#x^(1/3) = 5 or x^(1/3) = 1#

#(x^(1/3))^3 = 5^3 or (x^(1/3))^3 = 1^3#

#x = 125 or 1#

Hopefully this helps!