Question : If $x^2+\frac{1}{x^2}=7$, then what is the value of $x^3+\frac{1}{x^3}$?
Option 1: 9
Option 2: 18
Option 3: 27
Option 4: 36
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Correct Answer: 18
Solution : $x^2+\frac{1}{x^2}=7$ Adding 2 on both sides, we get, ⇒ $(x+\frac{1}{x})^2 - 2 × x × \frac{1}{x} = 7+2$ ⇒ $(x+\frac{1}{x})^2 = 9$ ⇒ $x+\frac{1}{x} = 3$ cubing on both sides, we get, ⇒ $(x+\frac{1}{x})^3 = 3^3$ ⇒ $x^3 + \frac{1}{x^3} + 3 × x × \frac{1}{x}(x+\frac{1}{x}) = 27$ ⇒ $x^3 + \frac{1}{x^3} + 3 × 3 = 27$ $\therefore x^3 + \frac{1}{x^3} =18$ Hence, the correct answer is 18.
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Question : If $x^{2}+\frac{1}{x^{2}}=1$, then, what is the value of $x^{48}+x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^{6}+1$?
Option 1: – 9
Option 2: 0
Option 3: 1
Option 4: 9
Question : If $\frac{x}{3}+\frac{3}{x}=1$, then the value of $x^3$ is:
Option 1: 1
Option 2: 27
Option 3: 0
Option 4: –27
Question : If $x+\frac{2}{x}=1$, then the value of $\frac{x^2+7x+2}{x^2+13x+2}$ is:
Option 1: $\frac{5}{7}$
Option 2: $\frac{3}{7}$
Option 3: $\frac{4}{7}$
Option 4: $\frac{2}{7}$
Question : If $\frac{x}{y}=\frac{4}{5}$, then the value of $(\frac{4}{7}+\frac{2y–x}{2y+x})$ is:
Option 1: $\frac{3}{7}$
Option 2: $1\frac{1}{7}$
Option 3: $1$
Option 4: $2$
Question : If $2:9::4:x$, then the value of $x$ is:
Option 1: 42
Option 3: 36
Option 4: 22
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