How do you solve #root3(2x^3 - 24) = -x# and find any extraneous solutions?

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Answer 1 · 2016-06-27T14:04:15

Cube both sides.

#(root(3)(2x^3 - 24))^3 = (-x)^3#

#2x^3 - 24 = -x^3#

#3x^3 = 24#

#x^3 = 8#

#x = 2#

Checking in the original equation:

#root(3)(2(2)^3 - 24 )=^? -2#

#root(3)(-8) = -2#

Therefore, out solution set is #{x = 2}#.

Hopefully this helps!