(৫)

$x + \frac{1}{x}= 4$ এবং x > 1

প্রদত্ত রাশি = $b{y}^2-(b^2+1)y+b$

$$\begin{gathered} =b{y}^2-b^{2}y-y+b \\ = by(y-b)-(y- b) \\ = (y - b)(by - 1) \end{gathered}$$

দেওয়া আছে,  $x + \frac{1}{x}= 4$

বা,$(x+\frac{1}{x}{)}^2=4^2$

বা,$(x-\frac{1}{x}{)}^2+4.x.\frac{1}{x}=16 [(a+b{)}^2=(a-b{)}^2+4ab]$

বা,$(x-\frac{1}{x}{)}^2=16-4$

বা,$(x-\frac{1}{x}{)}^2=12$

বা,$x - \frac{1}{x} =\sqrt{12}$

বা, $= \sqrt{4\times 3} =2 \sqrt{3}$

বামপক্ষ=

$$\begin{gathered} =x^3-\frac{1}{x^3}=(x-\frac{1}{x}{)}^3+3.x.\frac{1}{x}(x-\frac{1}{x}) \\ =(2\sqrt{3{)}^3}+3.2\sqrt{3} \\ =8\times 3\sqrt{3}+6\sqrt{3} \\ =24\sqrt{3}+6\sqrt{3} \\ =30\sqrt{3} \end{gathered}$$

= ডানপক্ষ

$x^3-\frac{1}{x^3}=30\sqrt{3}$  (প্রমাণিত)

দেওয়া আছে,  

$$\begin{gathered} x+\frac{1}{x}=4 \\ {x}^3+\frac{1}{x^3}=(x+\frac{1}{x}{)}^3-3.x.\frac{1}{x}(x+\frac{1}{x}) \\ =4^3-3\times 4=64-12=52 \end{gathered}$$

আবার, $x+\frac{1}{x}=4$

বা, ${\left(x+\frac{1}{x}\right)}^2=4^2$  [বর্গ করে]

বা, $x^2+2.x.\frac{1}{x}+\frac{1}{x^2}=16$

বা,$x^2+\frac{1}{x^2}=16-2$

$x^2+\frac{1}{x^2}=14$

এখন,  $(x^2+\frac{1}{x})(x^3+\frac{1}{x^3})=x^5+x+\frac{1}{x}+\frac{1}{x^5}$

$x^5+\frac{1}{x^5}+(x+\frac{1}{x})=(x^2+\frac{1}{x^2})(x^3+\frac{1}{x^3})$

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

= 14 $\times$52 - 4

=728-4= 724

$x+\frac{1}{x^5}=724$