Question
If $\text{x}=2+\sqrt{3},$ find the value of $\text{x}^3+\frac{1}{\text{x}^3}.$
✓
Answer
$\text{x}=2+\sqrt{3}$$\Rightarrow\frac{1}{\text{x}}=\frac{1}{2+\sqrt{3}}$
$=\frac{1}{2+\sqrt{3}}\times\frac{2-\sqrt{3}}{2-\sqrt{3}}$
$=\frac{2-\sqrt{3}}{2^2-\big(\sqrt{3}\big)^2}$
$=\frac{2-\sqrt{3}}{4-3}$
$=2-\sqrt{3}$
$\Rightarrow\text{x}+\frac{1}{\text{x}}=2+\sqrt{3}+2-\sqrt{3}=4$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=4^3$
$\Rightarrow\Big(\text{x}^3+\frac{1}{\text{x}^3}\Big)+3\times\text{x}\times\frac{1}{\text{x}}\Big(\text{x}+\frac{1}{\text{x}}\Big)=64$
$\Rightarrow\Big(\text{x}^3+\frac{1}{\text{x}^3}\Big)+3\times4=64$
$\Rightarrow\Big(\text{x}^3+\frac{1}{\text{x}^3}\Big)+12=64$
$\Rightarrow\Big(\text{x}^3+\frac{1}{\text{x}^3}\Big)=52$
$=\frac{1}{2+\sqrt{3}}\times\frac{2-\sqrt{3}}{2-\sqrt{3}}$
$=\frac{2-\sqrt{3}}{2^2-\big(\sqrt{3}\big)^2}$
$=\frac{2-\sqrt{3}}{4-3}$
$=2-\sqrt{3}$
$\Rightarrow\text{x}+\frac{1}{\text{x}}=2+\sqrt{3}+2-\sqrt{3}=4$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=4^3$
$\Rightarrow\Big(\text{x}^3+\frac{1}{\text{x}^3}\Big)+3\times\text{x}\times\frac{1}{\text{x}}\Big(\text{x}+\frac{1}{\text{x}}\Big)=64$
$\Rightarrow\Big(\text{x}^3+\frac{1}{\text{x}^3}\Big)+3\times4=64$
$\Rightarrow\Big(\text{x}^3+\frac{1}{\text{x}^3}\Big)+12=64$
$\Rightarrow\Big(\text{x}^3+\frac{1}{\text{x}^3}\Big)=52$
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