Question

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

• A. $6$
• B. $4$
• C. $5$
• D. $3$

Answer 1

The correct option is C $5$

$x=\frac{1}{2-\sqrt{3}}$

$x=\frac{1}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}}=\frac{2+\sqrt{3}}{4-3}=2+\sqrt{3}$

$\Rightarrow x=2+\sqrt{3}$

$\Rightarrow x-2=\sqrt{3}$

$\Rightarrow x^3={(2+\sqrt{3})}^3$

$\Rightarrow x^3=8+3\sqrt{3}+6\sqrt{3}(2+\sqrt{3})=8+3\sqrt{3}+12\sqrt{3}+18=26+15\sqrt{3}$

$\Rightarrow 3(x-1)(x+2)=3(2+\sqrt{3}-1)(2+\sqrt{3}+2)$

$=3(1+\sqrt{3})(4+\sqrt{3})$

$=3(4+\sqrt{3}+4\sqrt{3}+3)=3(7+5\sqrt{3})=21+15\sqrt{3}$

$\therefore x^3-3(x-1)(x+2)=26+15\sqrt{3}-21-15\sqrt{3}=5$