[tex]\dfrac{\sqrt[3]{x+h}-\sqrt[3]x}h\times\dfrac{\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}}{\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}}=\dfrac{(\sqrt[3]{x+h})^3-(\sqrt[3]x)^3}\cdots[/tex]
[tex]=\dfrac{x+h-x}\cdots=\dfrac h\cdots[/tex]
The [tex]h[/tex]s then cancel, leaving you with the [tex]\cdots=\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}[/tex] term.
If it's not clear what I did above, consider the substitution [tex]a=\sqrt[3]{x+h}[/tex] and [tex]b=\sqrt[3]x[/tex]. Then
[tex]a^3-b^3=(a-b)(a^2+ab+b^2)[/tex]
[tex]\implies\dfrac{a-b}h=\dfrac{a-b}h\times\dfrac{a^2+ab+b^2}{a^2+ab+b^2}=\dfrac{a^3-b^3}{h(a^2+ab+b^2)}[/tex]
