Question

If $x=\frac{5-\sqrt{21}}{2}$ find $(x+\frac{1}{x})(x^2+\frac{1}{x^2})(x^3+\frac{1}{x^3})$

Answer 1

$x=\frac{5-\sqrt{21}}{2}$

$\frac{1}{x}=\frac{2}{5-\sqrt{21}}$

On stationalising denominator

$\frac{1}{x}=\frac{2}{5.\sqrt{21}}\times \frac{5+\sqrt{21}}{5+\sqrt{21}}$

$\frac{1}{x}=\frac{2(5+\sqrt{21}}{25-21}=\frac{5+\sqrt{21}}{2}$

$x+\frac{1}{x}=\frac{5-\sqrt{21}}{2}+\frac{5+\sqrt{21}}{2}$

$\Rightarrow 5$

Now calculate $x^2+\frac{1}{x^2}$

$(x+\frac{1}{x})=5$ square on both sides

$x^2+\frac{1}{x^2}+2=25$

$x^2+\frac{1}{x^2}=23$

$\Rightarrow x^3+\frac{1}{x^3}\Rightarrow {(x+\frac{1}{x})}^3=(5{)}^3$

$=x^3+\frac{1}{x^3}+3(x+\frac{1}{x})=125$

$=x^3+\frac{1}{x^3}=110$

Hence $(x+\frac{1}{x})(x^2+\frac{1}{x^2})+(x^3+\frac{1}{x^3})=\mathrm{5.23.110}=12,650$.