text: 8x^2(x^2-4)^(1/3)+(x^2-4)^(4/3)=0 find all solutions of the equation - check for  extraneous solutions - can u answer from the book: x= +-2 and +-(2/3) please explain all your steps...show all work i wish i knew how to do this...i really need help with this problem (will be on my precal test in a few hours [around 5]!) thanks! will award those who help! you can split up the (x^2-4)^(4/3) into ((x^2-4)^(1/3))*(x^2-4), because you are adding the exponents when you multiply(1/3+1=4/3).  then you can factor out the (x^2-4)^(1/3) and come up with: ((x^2-4)^(1/3))*(8x^2+(x^2-4)) which simplifies to: ((x^2-4)^(1/3))*(9x^2-4)=0  that means two things: 9x^2-4=0 and (x^2-4)^(1/3)=0  take the first one, and you get the answer +-2/3 take the second one, cube both sides, and get x^2-4=0, which yields +-2.  plug all four of those answers into x to check for extraneous roots.  all four work, so there are no extraneous answers.  hope that helps!  tell me if something doesn't make sense.  edit: to the poster below me, ^ means "raised to the power of", and you have to hold your mouse over the equation in his post to get the whole thing.