Question : If $x=(\sqrt[3]{7})^{3}+3$, then the value of $x^3–9x^2+27x–34$ is:
Option 1: 0
Option 2: 1
Option 3: 2
Option 4: –1
Correct Answer: 0
Solution : Given: $(x–3)^{3}=(\sqrt[3]{7})^{3}$ $x^{3}–27–3(3x)(x–3) = 7$ $x^{3}–27–9x^{2}+27x = 7$ Subtracting 7 on both sides, we get, $x^{3}–27–9x^{2}+27x–7= 7–7$ $x^{3}–9x^{2}+27x–34 = 0$ Hence, the correct answer is 0.
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Question : If $a= \frac{\sqrt{x+2}+\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-2}}$, then the value of $(a^{2}-ax)$ is:
Option 1: 1
Option 2: 2
Option 3: –1
Option 4: 0
Question : If $x^2-\sqrt{7} x+1=0$, then $\left(x^3+x^{-3}\right)=?$
Option 1: $4 \sqrt{7}$
Option 2: $3 \sqrt{7}$
Option 3: $10 \sqrt{7}$
Option 4: $7 \sqrt{7}$
Question : If $x^2+\frac{1}{x^2}=\frac{7}{4}$ for $x>0$; then what is the value of $x+\frac{1}{x}$?
Option 1: $2$
Option 2: $\frac{\sqrt{15}}{2}$
Option 3: $\sqrt{5}$
Option 4: $\sqrt{3}$
Question : If $x^2+\frac{1}{x^2}=\frac{7}{4}$ for $x>0$, what is the value of $(x^3+\frac{1}{x^3})$?
Option 1: $\frac{3\sqrt{3}}{5}$
Option 2: $\frac{3\sqrt{15}}{5}$
Option 3: $\frac{3\sqrt{15}}{8}$
Option 4: $\frac{3\sqrt{5}}{8}$
Question : If $x^2-9x+1=0$, what is the value of $(x^3+\frac{1}{x^3})$?
Option 1: $54$
Option 2: $108$
Option 3: $702$
Option 4: $810$
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