সৃজনশীলঃ ‘x=5+23,y=5−2.…’ ফৌজদারহাট ক্যাডেট কলেজ ২০২৬ – পরিমিতি, গণিত (এসএসসি)

দেওয়া আছ, $x=\frac{\sqrt{5}+\sqrt{2}}{3}$
বা, $3x=\sqrt{5}+\sqrt{2}$
এবং $y=\sqrt{5}-\sqrt{2}$

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

এবং $3x-y=\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}=2\sqrt{2}$
প্রদত্ত রাশি $=3xy(9x^2+y^2)$ 

$\begin{array}{cc}& =\frac{1}{8}\cdot 4\cdot 3x\cdot y\cdot 2(9x^2+y^2)\\ & =\frac{1}{8}\left\{\right(3x+y{)}^2-(3x-y{)}^2\}\left\{\right(3x+y{)}^2+(3x-y{)}^2\}\\ & =\frac{1}{8}\left\{\right(2\sqrt{5}{)}^2-\left(2\sqrt{2}{)}^2\right\}\left\{\right(2\sqrt{5}{)}^2+\left(2\sqrt{2}{)}^2\right\}\\ & =\frac{1}{8}(20-8)(20+8)=\frac{1}{8}\times 12\times 28=42\end{array}$ 

নির্ণেয় মান: 42. 

দেওয়া আছে, $y=\sqrt{5}-\sqrt{2}$ 

$\therefore \frac{1}{y}=\frac{1}{\sqrt{5}-\sqrt{2}}$ 

বা, $\frac{3}{y}=\frac{3}{\sqrt{5}-\sqrt{2}}=\frac{3(\sqrt{5}+\sqrt{2})}{(\sqrt{5}-\sqrt{2})(\sqrt{5}+2)}=\frac{3(\sqrt{5}+\sqrt{2})}{(\sqrt{5}{)}^2-(\sqrt{2}{)}^2}$ 

$\begin{array}{c}=\frac{3(\sqrt{5}+\sqrt{2})}{5-2}=\frac{3(\sqrt{5}+\sqrt{2})}{3}=\sqrt{5}+\sqrt{2}\\ \therefore \frac{3}{y}-y=\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}-2\sqrt{2}.\end{array}$ 

বামপক্ষ $=\frac{27}{y^3}-y^3$ 

$\begin{array}{cc}& ={\left(\frac{3}{y}\right)}^3-y^3={(\frac{3}{y}-y)}^3+3\cdot \frac{3}{y}\cdot y(\frac{3}{y}-y)\\ & =(2\sqrt{2}{)}^3+9\times 2\sqrt{2}=16\sqrt{2}+18\sqrt{2}-34\sqrt{2}\end{array}$ 

= ডানপক্ষ

$\therefore \frac{27}{y^3}-y^3=34\sqrt{2}$. (প্রমাণিত)