$x^2+\frac{1}{x^2}=10$ একটি বীজগাণিতিক রাশি।  

দেওয়া আছে, $x^2+\frac{1}{x^2}=10$

এখন, ${\left(x+\frac{1}{x}\right)}^2=x^2+\frac{1}{x^2}+2.x\frac{1}{x}$

                                $=10+2=12$

   $$\begin{gathered} \therefore x +\frac{ 1}{x}= \sqrt{12} \\ =2\sqrt{3} \end{gathered}$$   নির্ণেয় মান : $2\sqrt{3}$

দেওয়া আছে, $x^2+\frac{1}{x^2}=10$  

ক-হতে প্রাপ্ত , $x + \frac{1}{x }=2\sqrt{3}$

এখন, ${\left(x-\frac{1}{x}\right)}^2=x^2+\frac{1}{x^2}-2.x\frac{1}{x}=10-2=8$

                                $\therefore x - \frac{1}{x}=\sqrt{8}=2\sqrt{2}$

                                   $x^2-\frac{1}{x^2}=\left(x+\frac{1}{x}\right) \left(x-\frac{1}{x}\right)=2\sqrt{3.} 2\sqrt{2}=4\sqrt{6}$

বামপক্ষ= $=\frac{x^8-1}{x^4}=x^4-\frac{1}{x^4}=(x^2{)}^2-{\left(\frac{1}{x^2}\right)}^2=\left(x^2+\frac{1}{x^{2 }}\right) \left( x^2-\frac{1}{x^2}\right)$

                    $=10\times 4\sqrt{6}=40\sqrt{6}$ = R.H.T

                        $\therefore \frac{x^8-1}{x^4}=40\sqrt{6}$ (প্রমাণিত)

ক-হতে প্রাপ্ত, $x + \frac{1}{x}= 2\sqrt{3}$

খ-হতে প্রাপ্ত, $x - \frac{1}{x}=2\sqrt{2}$

এবং, $x^2-\frac{1}{x^2}=4\sqrt{6}$

এখন, $x^3+\frac{1}{x^3}={\left(x+\frac{1}{x}\right)}^3-3.x,\frac{1}{x}\left(x+\frac{1}{x}\right)$

                              ${\left(2\sqrt{3}\right)}^3-3.2\sqrt{3}=24\sqrt{3}-6\sqrt{3}=18\sqrt{3}$

প্রদত্ত রাশি= $x^5-\frac{1}{x^5}=\left(x^3+\frac{1}{x^3}\right) \left(x^2-\frac{1}{x^2}\right)+\left(x-\frac{1}{x}\right)$

                                      $=18\sqrt{3}. 4\sqrt{6}+2\sqrt{2}=72\sqrt{18}+2\sqrt{2}$

                                         $=72.3\sqrt{2}+2\sqrt{2}=216\sqrt{2}+2\sqrt{2}=218\sqrt{2}$

        নির্ণেয় মান: $218\sqrt{2}.$