There have now been so many posts that I cannot remember what you have shown and what not. Let's take it in pieces.
[tex]\displaystyle f(x) = (2x)^{1/3} = \sqrt[3]{2} * x^{1/3} \implies \int \ f(x) \ dx = \dfrac{3 \sqrt[3]{2} * x^{4/3}}{4} + C_1 = F(x).[/tex]
[tex]\displaystyle g(x) = \dfrac{x^2}{8} \implies \int \ g(x) \ dx = \dfrac{x^3}{24} + C_2 = G(x).[/tex]
Obviously f(0) = 0 = g(0).
[tex]x > 0 \implies f(x) \div g(x) = 8 \sqrt[3]{2} * \dfrac{1}{x^{5/3}}.[/tex]
Now it is obvious by inspection that
[tex] 0 < x \le 1 \implies f(x) \div g(x) > 1 \because \dfrac{1}{x^{5/3}} \ge 1 \text { and } 8\sqrt[3]{2} > 1.[/tex]
It is also obvious by inspection that
[tex]f(x) \div g(x)[/tex] is monotonically falling as x increases: the numerator is a constant whereas the denominator is increasing as x does.
So the question is where (if anywhere) does that quotient = 1.
[tex]f(x) \div g(x) = 1 \implies x^{5/3} = 8\sqrt[3]{2} = x^{5/3} \implies x^5 = 1024 \implies x = 4.[/tex]
So the area sought =
[tex](\{F(4) - F(0)\} - \{G(4) - G(0)\}) + (\{G(6) - G(4)\} - \{F(6) - F(4)\}) = what?[/tex]
By the way, I get the same answer your book does. I have no idea where your algebra or arithmetic went astray.
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