Quote:
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Originally Posted by Pip
Lesson: "Raise to the power of" does not behave like "multiply with".
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I'm assuming here you mean that "rooting x does not mean raise to the power of x as it's inverse operation"?
Well it does really. The problem comes from that (7x-2)^(1/3) + (7x+5)^(1/3) does
not equal [(7x-2)+ (7x+5)]^1/3.
For instance if the problem
was [(7x-2)+ (7x+5)]^1/3 = 3 then we could cube both sides and it would work.
Solving it we get 12/7ths like has been shown before, and when we plug it in: [(7(12/7)-2)+ (7(12/7)+5)]^1/3 = 3
cubing it
(12-2)+ (12+5) = 27 and bingo it works.
To recap, (7x-2)^(1/3) + (7x+5)^(1/3) does
not equal [(7x-2)+ (7x+5)]^1/3.
In other words when you cube both sides of (7x-2)^(1/3) + (7x+5)^(1/3) = 3 you are not really getting rid of the two ^1/3s you only think you are.