Question

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

• A. 0
• B. 1
• C. 2
• D. -1

Answer 1

Answer: (A)

0

We will make use of the following identities to solve the question :

${(a+b)}^2+{(a-b)}^2=2(a^2+b^2)⟶\left(1\right)$

${(a+b)}^3+{(a-b)}^3=2a(a^2+{3b}^2)⟶\left(2\right)$

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

$\therefore$ $\frac{1}{x}=\frac{2}{5-\sqrt{21}}=\frac{2(5+\sqrt{21})}{(5-\sqrt{21})(5+\sqrt{21})}=\frac{2(5+\sqrt{21})}{4}$

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

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

Similarly, $x^2+\frac{1}{x^2}={\left(\frac{5-\sqrt{21}}{2}\right)}^2+{\left(\frac{5+\sqrt{21}}{2}\right)}^2$

Using equation $\left(1\right)$ we can easily write

$x^2+\frac{1}{x^2}=2\times (\frac{25}{4}+\frac{21}{4})=23$

Also, $x^3+\frac{1}{x^3}={\left(\frac{5-\sqrt{21}}{2}\right)}^3+{\left(\frac{5+\sqrt{21}}{2}\right)}^3$

Using equation $\left(2\right)$,

$x^3+\frac{1}{x^3}=2\times \frac{5}{2}\times (\frac{25}{4}+\frac{63}{4})=110$

$\therefore$ $(x^3+\frac{1}{x^3})-5(x^2+\frac{1}{x^2})+(x+\frac{1}{x})=110-5\times 23+5=115-115=0$

Correct answer : Option A.