Question
If $\text{x}=2+\sqrt3,$ find the value of $\text{x}^3+\frac{1}{\text{x}^3}.$
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Answer
We know that $\text{x}^3+\frac{1}{\text{x}^3}=\Big(\text{x}+\frac{1}{\text{x}}\Big)\Big(\text{x}^2-1+\frac{1}{\text{x}^2}\Big).$ we have to find the value of $\text{x}^3+\frac{1}{\text{x}^3}.$ As $\text{x}=2+\sqrt3$ therefore,$\frac{1}{\text{x}}=\frac{1}{2+\sqrt3}$
We know that rationalization factor for $2+\sqrt3$ is $2-\sqrt3.$ We will multiply numerator and denominator of the given expression $\frac{1}{2+\sqrt3}$ by $2-\sqrt3,$ to get.$\frac{1}{\text{x}}=\frac{1}{2+\sqrt3}\times\frac{2-\sqrt3}{2-\sqrt3}$
$=\frac{2-\sqrt3}{(2)^2-\big(\sqrt3\big)^2}$
$=\frac{2-\sqrt3}{4-3}$
$=2-\sqrt3$
Putting the value of x and $\frac{1}{\text{x}},$ we get$\text{x}^3+\frac{1}{\text{x}^3}=\big(2+\sqrt3+2-\sqrt3\big)\Big(\big(2+\sqrt3\big)^2-1+\big(2-\sqrt3\big)^2\Big)$
$=4\Big(2^2+\big(\sqrt3\big)^2+2\times2\times\sqrt3-1+2^2+\big(\sqrt3\big)^2-2\times2\times\sqrt3\Big)$
$=4\big(4+3+4\sqrt3-1+4+3-4\sqrt3\big)$
$=52$
Hence the value of the given expression 52.
We know that rationalization factor for $2+\sqrt3$ is $2-\sqrt3.$ We will multiply numerator and denominator of the given expression $\frac{1}{2+\sqrt3}$ by $2-\sqrt3,$ to get.$\frac{1}{\text{x}}=\frac{1}{2+\sqrt3}\times\frac{2-\sqrt3}{2-\sqrt3}$
$=\frac{2-\sqrt3}{(2)^2-\big(\sqrt3\big)^2}$
$=\frac{2-\sqrt3}{4-3}$
$=2-\sqrt3$
Putting the value of x and $\frac{1}{\text{x}},$ we get$\text{x}^3+\frac{1}{\text{x}^3}=\big(2+\sqrt3+2-\sqrt3\big)\Big(\big(2+\sqrt3\big)^2-1+\big(2-\sqrt3\big)^2\Big)$
$=4\Big(2^2+\big(\sqrt3\big)^2+2\times2\times\sqrt3-1+2^2+\big(\sqrt3\big)^2-2\times2\times\sqrt3\Big)$
$=4\big(4+3+4\sqrt3-1+4+3-4\sqrt3\big)$
$=52$
Hence the value of the given expression 52.
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