Question : If $x=a+\frac{1}{a}$ and $y=a-\frac{1}{a}$, then the value of $x^4+y^4-2x^2y^2$ is:
Option 1: 4
Option 2: 8
Option 3: 16
Option 4: 64
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Correct Answer: 16
Solution : Given: $x=a+\frac{1}{a}$, $y=a-\frac{1}{a}$ Now, $x^2=(a+\frac{1}{a})^2=a^2+\frac{1}{a^2}+2$ And $y^2=(a-\frac{1}{a})^2=a^2+\frac{1}{a^2}-2$ Now, we know, $x^4+y^4-2x^2y^2=(x^2-y^2)^2$ $⇒x^4+y^4-2x^2y^2= [(a^2+\frac{1}{a^2}+2)-(a^2+\frac{1}{a^2}-2)]^2$ $\therefore x^4+y^4-2x^2y^2=(2+2)^2=16$ Hence, the correct answer is 16.
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Question : If $x=\sqrt{a}+\frac{1}{\sqrt{a}}$ and $y=\sqrt{a}-\frac{1}{\sqrt{a}}, (a>0),$ then the value of $(x^{4}+y^{4}-2x^{2}y^{2})$ is:
Question : If $(x-\frac{1}{3})^2+(y-4)^2=0$, then what is the value of $\frac{y+x}{y-x}$?
Question : If $2x-\frac{2}{x}=1(x \neq 0)$, then the the value of $(x^3-\frac{1}{x^3})$ is:
Question : If $2x+\frac{2}{x}=3$, then the value of $x^{3}+\frac{1}{x^{3}}+2$ is:
Question : When $2x+\frac{2}{x}=3$, then the value of ($x^3+\frac{1}{x^3}+2)$ is:
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