Question

If $x=\sqrt[3]{328},y=\sqrt[3]{327}$, then the value of $x+y-\frac{1}{x^2+xy+y^2}$ is

• A. $8$
• B. $7$
• C. $6$
• D. $5$

Answer 1

The correct option is C $6$

$x=\sqrt[3]{328},y=\sqrt[3]{327}$

Then

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

$=x+y-\frac{x-y}{(x-y)(x^2+xy+y^2)}$

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

$=x+y-\frac{x-y}{28+27}=x+y-(x-y)$

$=2y$

$=2\times \sqrt[3]{327}$

$=2\times 3=6$

$x+y-\frac{1}{x^2+xy+yz}=6$