If $x = \sqrt[3]{28}$ and $y = \sqrt[3]{27}$, find the value of $x + y - \frac{1}{x^2+xy+y^2}$.

Since $y = \sqrt[3]{27}$ and $27 = 3^3$:

$y = \sqrt[3]{3^3}$

$y = 3$

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The expression $x^2 + xy + y^2$ in the denominator connects to the factorization:

$x^3 - y^3 = (x-y)(x^2+xy+y^2)$

Therefore:

$x^2 + xy + y^2 = \dfrac{x^3 - y^3}{x-y}$

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Since $x = \sqrt[3]{28}$ and $y = \sqrt[3]{27}$:

$x^3 = (\sqrt[3]{28})^3 = 28$

$y^3 = (\sqrt[3]{27})^3 = 27$

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Substituting:

$x^2 + xy + y^2 = \dfrac{x^3 - y^3}{x-y}$

$x^2 + xy + y^2 = \dfrac{28 - 27}{x-y}$

$x^2 + xy + y^2 = \dfrac{1}{x-y}$

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The original expression becomes:

$x + y - \dfrac{1}{x^2+xy+y^2} = x + y - \dfrac{1}{\dfrac{1}{x-y}}$

$= x + y - (x-y)$

$= x + y - x + y$

$= 2y$

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Since $y = 3$:

$2y = 2 \times 3 = 6$