If left(x+frac{1}{x}right)^{2}=3,then the val | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $\left(x+\frac{1}{x}\right)^{2}=3$.Taking the square root on both sides,we get $x+\frac{1}{x}=\sqrt{3}$.Cubing both sides,we get $\left(x+\f

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Given $\left(x+\frac{1}{x}\right)^{2}=3$.Taking the square root on both sides,we get $x+\frac{1}{x}=\sqrt{3}$.Cubing both sides,we get $\left(x+\frac{1}{x}\right)^{3} = (\sqrt{3})^{3}$.Using the identity $(a+b)^{3} = a^{3}+b^{3}+3ab(a+b)$,we have $x^{3}+\frac{1}{x^{3}}+3\left(x+\frac{1}{x}\right) = 3\sqrt{3}$.Substituting $x+\frac{1}{x}=\sqrt{3}$,we get $x^{3}+\frac{1}{x^{3}}+3(\sqrt{3}) = 3\sqrt{3}$.This simplifies to $x^{3}+\frac{1}{x^{3}} = 0$,which implies $x^{6}+1 = 0$.Now,the given expression is $x^{206}+x^{200}+x^{90}+x^{84}+x^{18}+x^{12}+x^{6}+1$.Grouping the terms,we get $x^{200}(x^{6}+1)+x^{84}(x^{6}+1)+x^{12}(x^{6}+1)+1(x^{6}+1)$.Since $x^{6}+1=0$,the entire expression becomes $x^{200}(0)+x^{84}(0)+x^{12}(0)+1(0) = 0$.

If $\left(x+\frac{1}{x}\right)^{2}=3$,then the value of $x^{206}+x^{200}+x^{90}+x^{84}+x^{18}+x^{12}+x^{6}+1$ is

Similar Questions

Evaluate $\frac{0.53 \times 0.53 - 2 \times 0.53 \times 0.41 + 0.41 \times 0.41}{0.53 - 0.41}$

What will come in place of the question mark (?) in the following question?
$335.01 \times 244.99 \div 55 = ?$

Which of the following is $TRUE$?
$I.$ $\sqrt[3]{11} > \sqrt{7} > \sqrt[4]{45}$
$II.$ $\sqrt{7} > \sqrt[3]{11} > \sqrt[4]{45}$
$III.$ $\sqrt{7} > \sqrt[4]{45} > \sqrt[3]{11}$
$IV.$ $\sqrt[4]{45} > \sqrt{7} > \sqrt[3]{11}$

The difference between two numbers is $9$ and the difference between their squares is $207$. The numbers are

If $x = (0.25)^{\frac{1}{2}}, y = (0.5)^{2}, z = (0.216)^{\frac{1}{3}},$ then

Vedclass Test Series

Exam Paper Generator

Online Exam Module