Question : If $x=\sqrt[3]{28}, y=\sqrt[3]{27}$, then the value of $x+y-\frac{1}{x^{2}+xy+y^{2}}$ is:
Option 1: 8
Option 2: 7
Option 3: 6
Option 4: 5
Correct Answer: 6
Solution :
Given: $x=\sqrt[3]{28}$ and $y=\sqrt[3]{27}.$
To find $x+y–\frac{1}{x^{2}+xy+y^{2}}$
Multiplying and dividing by $(x–y)$
$=x+y–\frac{(x–y)}{(x–y)(x^{2}+xy+y^{2})}$
$=x+y–\frac{(x–y)}{(x^{3}–y^{3})}$
$=x+y–\frac{(x–y)}{(28–27)}$
$=x+y–x+y$
$=2y$
$=2×\sqrt[3]{27}$
$=2×3 = 6$
Thus, $x+y-\frac{1}{x^{2}+xy+y^{2}}=6$
Hence, the correct answer is 6.
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