Question : If $\frac{x^8+1}{x^4}=14$, then the value of $\frac{x^{12}+1}{x^6}$ is:
Option 1: 16
Option 2: 14
Option 3: 52
Option 4: 64
Correct Answer: 52
Solution : $\frac{x^8+1}{x^4}=14$ $⇒x^4+\frac{1}{x^4}=14$ $⇒x^4+\frac{1}{x^4}+2=14+2$ $⇒(x^2+\frac{1}{x^2})^2=(4)^2$ $⇒x^2+\frac{1}{x^2}=4$ Now, $\frac{x^{12}+1}{x^6}$ $=x^6+\frac{1}{x^6}$ $=(x^2)^3+\frac{1}{(x^2)^3}$ $=(x^2+\frac{1}{x^2})^3-3×x^2×\frac{1}{x^2}(x^2+\frac{1}{x^2})$ $=4^3-3(4)$ $=64-12=52$ Hence, the correct answer is 52.
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Question : What is the simplified value of $(x^{32}+\frac{1}{x^{32}})(x^{8}+\frac{1}{x^{8}})(x-\frac{1}{x})(x^{16}+\frac{1}{x^{16}})(x+\frac{1}{x})(x^{4}+\frac{1}{x^{4}})?$
Option 1: $(x^{64}+\frac{1}{x^{64}})$
Option 2: $\frac{(x^{64}-\frac{1}{x^{64}})}{(x^2+\frac{1}{x^2})}$
Option 3: $\frac{(x^{64}-\frac{1}{x^{64}})}{(x+\frac{1}{x})}$
Option 4: $\frac{(x^{32}-\frac{1}{x^{32}})}{(x+\frac{1}{x})}$
Question : If $\frac{x^{2}-x+1}{x^{2}+x+1}=\frac{2}{3}$, then the value of $\left (x+\frac{1}{x} \right)$ is:
Option 1: 4
Option 2: 5
Option 3: 6
Option 4: 8
Question : If $x^4+x^{-4}=194, x>0$, then the value of $x+\frac{1}{x}$ is:
Option 1: 14
Option 2: 6
Option 3: 4
Question : What is the value of $\frac{x^2-x-6}{x^2+x-12}÷\frac{x^2+5x+6}{x^2+7x+12}$?
Option 1: $1$
Option 2: $\frac{(x-3)}{(x+3)}$
Option 3: $\frac{(x+4)}{(x-3)}$
Option 4: $\frac{(x-3)}{(x+4)}$
Question : If $x$4 + $x$ -4 = 194, x > 0, then what is the value of $x+\frac{1}{x}+2$?
Option 1: 8
Option 4: 4
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