Question

If ${(x+\frac{1}{x})}^2=3$. Find value of $x^{18}+x^{12}+x^6+1$.

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

We have,

$x+\frac{1}{x}=\sqrt{3}$

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

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

$\Rightarrow x^3+\frac{1}{x^3}+3\left(\sqrt{3}\right)=3\sqrt{3}$

$\Rightarrow x^3+\frac{1}{x^3}=0$

$\Rightarrow x^3=\frac{-1}{x^3}\Rightarrow x^6=-1$

squaring $(x^6{)}^2=(-1{)}^2$

$\Rightarrow x^{12}=1$

Cubing $\Rightarrow (x^6{)}^3=(-1{)}^3=-1$

$\Rightarrow x^{18}=-1$

$x^{18}+x^{12}+x^6+1=-1+1-1+1=0$