Question
Show that: $x^3+\frac{1}{x^3}=52$, if $\mathrm{x}=2+\sqrt{3}$
✓
Answer
Given: $x=2+\sqrt{ } 3$
$\frac{1}{x}=\frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}}$
$=\frac{2-\sqrt{3}}{(2)^2-(\sqrt{3})^2}$
$=\frac{2-\sqrt{3}}{4-3}$
$=2-\sqrt{3}$
Now,
$x+\frac{1}{2}=2+\sqrt{3}+2-\sqrt{3}$
$x+\frac{1}{x}=2+2$
$x+\frac{1}{x}=4$
$\therefore x^3+\frac{1}{x^3}$
$=\left(x+\frac{1}{x}\right)^3-3 \cdot \not x \cdot \frac{1}{\not x}\left(x+\frac{1}{x}\right)$
$=(4)^3-3 \times 4$
$=64-12$
$=52$
$\frac{1}{x}=\frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}}$
$=\frac{2-\sqrt{3}}{(2)^2-(\sqrt{3})^2}$
$=\frac{2-\sqrt{3}}{4-3}$
$=2-\sqrt{3}$
Now,
$x+\frac{1}{2}=2+\sqrt{3}+2-\sqrt{3}$
$x+\frac{1}{x}=2+2$
$x+\frac{1}{x}=4$
$\therefore x^3+\frac{1}{x^3}$
$=\left(x+\frac{1}{x}\right)^3-3 \cdot \not x \cdot \frac{1}{\not x}\left(x+\frac{1}{x}\right)$
$=(4)^3-3 \times 4$
$=64-12$
$=52$
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