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

Given : $$x=\sqrt[3]{28},y=\sqrt[3]{27}$$

=> $$x^3=28$$ and $$y^ 3=27$$ ----------(i)

=> $$y=3$$ -----------(ii)

To find : $$x+y-\frac{1}{x^2+xy+y^2}$$

= $$(x+y)-(\frac{(x-y)}{(x-y)(x^2+xy+y^2)})$$     [Multiply and divide by $$(x-y)$$]

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

= $$(x+y)-(\frac{x-y}{x^3-y^3})$$

= $$(x+y)-(\frac{x-y}{28-27})$$     [Using (i)]

= $$(x+y)-(x-y)=2y$$

= $$2 \times 3=6$$     [Using (ii)]

=> Ans - (C)