প্রদত্ত রাশি $=a^3+9b^3+(a+b{)}^3$

                    $=a^3+8b^3+(a+b{)}^3+b^3$

                    $=a^3+(2b{)}^3+(a+b{)}^3+b^3$

                    $=(a+2b)(a^2-a.2b+4b^2)+(a+b+b)\left\{\right(a+b{)}^2-(a+b)b+b^2\}$

                  $=(a+2b)(a^2-2ab+4b^2)+(a+2b)(a^2+2ab+b^2-ab-b^2+b^2)$

                   $=(a+2b)(a^2-2ab+4b^2)+(a+2b)(a^2+ab+b^2)$

                 $=(a+2b)(a^2-2ab+4b^2+a^2+ab+b^2)$

                  $=(a+2b)(2a^2-ab+5b^2)$

দেওয়া আছে, $x^2-\sqrt{3}x=1$

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

বা, $x - \sqrt{3} = \frac{1}{x}$

বা, $x -\frac{ 1}{x}=\sqrt{3}$

বা, ${x-\frac{1}{x}}^2={\left(\sqrt{3}\right)}^2$

বা, $x^2-2.x.\frac{1}{x}+\frac{1}{x^2}=3$

বা, $x^2+\frac{1}{x^2}=3+2=5$

বা, $x^2+\frac{1}{x^2}=5^2$

বা $x^4+2.x^2.\frac{1}{x^2}+\frac{1}{x^4}=25$

বা, $x^4+\frac{1}{x^4}=25-2$

$\therefore x^4+\frac{1}{x^4}=23$

$\therefore \left(x^2+\frac{1}{x^2}\right):\left(x^4+\frac{1}{x^4}\right)=5:23$ (প্রমাণিত)

খ' হতে পাই, $x^2+\frac{1}{x^2}=5$

   প্রদত্ত রাশি$=x^6+\frac{1}{x^6}$

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

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

                        $$\begin{gathered} =5^3-3\times 5 \\ =125-15=110 \end{gathered}$$

প্রদত্ত রাশি $=x^3-9y^3+(x-y{)}^3$

                   $=x^3-8y^3+(x-y{)}^3-y^3$

                  $=(x{)}^3-.(2y{)}^3+(x-y{)}^3-(y{)}^3$

                  $=(x-2y)(x^2+x0.2y+4y^2)+(x-y-y) (x-y{)}^2+(x-y).y+y^2$

                  $=(x-2y)(x^2+2xy+4y^2)+(x-2y)(x^2-2xy+y^2+xy-y^2+y^2)$

                   $=(x-2y)(x^2+2xy+4y^2)+(x-2y)(x^2-xy+y^2)$

                    $=(x-2y)(x^2+2xy+4y^2+x^2-xy+y^2)$

                   $=(x-2y)(2x^2+xy+5y^2)$

দেওয়া আছে, $S=\left\{\right(x,y):x\in A,y\in A$ এবং $y=\sqrt{x}\}$

যেখানে, $A=\{-9,-3,-1,0,1,3,9\}$

S এর বর্ণিত শর্ত থেকে পাই ,$y = \sqrt{x}$

এখন, প্রত্যেক $x\in A$ এর জন্য $y = \sqrt{x}$ এর মান নির্ণয় করি,

যেহেতু $\sqrt{9},\sqrt{-3},\sqrt{-1},\sqrt{3}\notin A$

সেহেতু $\left(-9,\sqrt{-9}\right),\left(-3,\sqrt{-3}\right)\left(-1,\sqrt{-1}\right),\left(3,\sqrt{3}\right)\notin S$

$\therefore S=\{ (0,0),(1,1),(9,3)$

$\therefore$ রেঞ্জ , $S=\{0,1,3\}$

দেওয়া আছে, $B=x^3+x^{-3}=x^3+\frac{1}{x^3}$

$B = 18$ হলে, $x^3+\frac{1}{x^3}=18$

বা, $\frac{x^6+1}{x^3}=18$

বা, $x^6+1=18x^3$

বা, $x^6-18x^3+1=0$

বা, $(x^3{)}^2-2.x^3.9+9^2-(9{)}^2+1=0$

বা, $(x^3-9{)}^2-81+1=0$

বা, $(x^3-9{)}^2=80$

বা, $x^3-9=\sqrt{80}=\sqrt{16\times 5}=4\sqrt{5}$

বা, $x^3=\frac{27}{8}+\frac{27\sqrt{5}}{8}+\frac{45}{8}+\frac{5\sqrt{5}}{8}$

বা, $x^3={\left(\frac{3}{2}\right)}^2+3{\left(\frac{3}{2}\right)}^2.\frac{\sqrt{5}}{2}+3.\frac{3}{2}.{\left(\frac{\sqrt{5}}{2}\right)}^2+{\left(\frac{\sqrt{5}}{2}\right)}^3$

বা, $x^3={\left(\frac{3}{2}+\frac{\sqrt{5}}{2}\right)}^3$

বা, $x=\frac{3}{2}+\frac{\sqrt{5}}{2}$ [ঘনমূল করে।]

$\therefore x = \frac{3 +\sqrt{5}}{2}$

নির্ণেয় মান: $x = \frac{3+\sqrt{5}}{2}$