দেখাও যে ,$f\left(2\right) = f\left(3\right)$

দেওয়া আছে, $f\left(x\right)=x^3-6x^2+11x-6$

এখন, $f\left(x\right) = 0$

বা, $x^3-6x^2+11x-6=0$

বা $x^3-x^2-5x^2+5x+6x-6=0$

বা $x^2(x-1)-5x(x-1)+6(x-1)=0$

বা, $(x-1)(x^2-5x+6)=0$

বা, $(x-1)(x^2-2x-3x+6)=0$

বা, $(x-1)\left\{x\right(x-2)-3(x-2\left)\right\}=0$

বা, $(x - 1)(x - 2)(x - 3) = 0$

হয়, $x - 1 = 0$

$\therefore x = 1$                      অথবা, $x - 2 = 0$                         অথবা, $x - 3 = 0$

                                                        $\therefore x = 2$                                             x = 3

অতএব, x এর মান 1 বা 2 বা 3 এরূ জন্য f(x) = 0 হবে।

দেওয়া আছে, $g\left(x\right)=\frac{1+x^2+x^4}{x^2}$

       $\therefore g\left(y^{-2}\right)=\frac{(1+(y^{-2}{)}^2+(y^{-2}{)}^4}{(y^{-2}{)}^2}$

      $=\frac{1+y^{-4}+y^{-8}}{y^{-4}}=\frac{1+{\displaystyle \frac{1}{y^4}}+{\displaystyle \frac{1}{y^8}}}{{\displaystyle \frac{1}{y^4}}}=\frac{{\displaystyle \frac{y^8+y^4+1}{y^8}}}{{\displaystyle \frac{1}{y^4}}}$

$=\frac{1+y^4+y^8}{y^8}\times \frac{y^4}{1}=\frac{1+y^4+y^8}{y^4}$

আবার, $g\left(y^2\right)=\frac{(1+(y^2{)}^2+\left(y^2{)}^4\right)}{(y^2{)}^2}=\frac{1+y^4+y^8}{y^4}$

$\therefore g\left(y^{-2}\right)=g\left(y^2\right)$  (দেখানো হলো)