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
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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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Question : If $x=\frac{\sqrt{5}+1}{\sqrt{5}-1}$ and $y=\frac{\sqrt{5}-1}{\sqrt{5}+1}$, then the value of $\frac{x^{2}+xy+y^{2}}{x^{2}-xy+y^{2}}$ is:
Option 1: $\frac{3}{4}$
Option 2: $\frac{4}{3}$
Option 3: $\frac{3}{5}$
Option 4: $\frac{5}{3}$
Question : If $x=\frac{\sqrt{5}-\sqrt{4}}{\sqrt{5}+\sqrt{4}}$ and $y=\frac{\sqrt{5}+\sqrt{4}}{\sqrt{5}-\sqrt{4}}$ then the value of $\frac{x^2-x y+y^2}{x^2+x y+y^2}=$?
Option 1: $\frac{361}{363}$
Option 2: $\frac{341}{343}$
Option 3: $\frac{384}{387}$
Option 4: $\frac{321}{323}$
Question : If $x^2+y^2=29$ and $xy=10$, where $x>0,y>0$ and $x>y$. Then the value of $\frac{x+y}{x-y}$ is:
Option 1: $- \frac{7}{3}$
Option 2: $\frac{7}{3}$
Option 3: $\frac{3}{7}$
Option 4: $-\frac{3}{7}$
Question : If $x^2-\sqrt{7} x+1=0$, then what is the value of $x^5+\frac{1}{x^5} ?$
Option 1: $19 \sqrt{7}$
Option 2: $21 \sqrt{7}$
Option 3: $25 \sqrt{7}$
Option 4: $27 \sqrt{7}$
Question : If $x+\left [\frac{1}{(x+7)}\right]=0$, what is the value of $x-\left [\frac{1}{(x+7)}\right]$?
Option 1: $3\sqrt{5}$
Option 2: $3\sqrt{5}-7$
Option 3: $3\sqrt{5}+7$
Option 4: $8$
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