উদাহরণ ৪৷
সমীকরণর একটি বীজ অন্যটর বর্গ হলে প্রমান করো যে, 
[H.S ‘82]
সমাধান: মনে করি সমকরণের একটি বীজ হল তাহলে অন্যটি হবে 
।
শর্তানুযায়ী
-br /-\end{array}.jpg "\begin{array}{l}<br />\alpha + {\alpha ^2} = - r \to \left( 1 \right)\\<br />\alpha \times {\alpha ^2} = - s \Rightarrow {\alpha ^3} = - s \to \left( 2 \right)<br />\end{array}")
(1) থেকে পাই
\alpha + {\alpha ^2} = - r\\
\Rightarrow \alpha \left( {1 + \alpha } \right) = - r\\
\Rightarrow {\alpha ^3}{\left( {1 + \alpha } \right)^3} = - {r^3}\\
\Rightarrow \left( { - s} \right){\left( {1 + \alpha } \right)^3} = - {r^3}\left[ {{\alpha ^3} = - s} \right]\\
\Rightarrow s{\left( {1 + \alpha } \right)^3} = {r^3}\\
\Rightarrow s\left( {{\alpha ^3} + 3{\alpha ^2} + 3\alpha + 1} \right) = {r^3}\\
\Rightarrow s\left( { - s + 3\alpha \left( {\alpha + 1} \right) + 1} \right) = {r^3}\\
\Rightarrow s\left( { - s - 3r + 1} \right) = {r^3}\\
\Rightarrow {r^3} + {s^2} + 3rs - s = 0\left( {proved} \right)
\end{array}" alt="\begin{array}{l}
\alpha + {\alpha ^2} = - r\\
\Rightarrow \alpha \left( {1 + \alpha } \right) = - r\\
\Rightarrow {\alpha ^3}{\left( {1 + \alpha } \right)^3} = - {r^3}\\
\Rightarrow \left( { - s} \right){\left( {1 + \alpha } \right)^3} = - {r^3}\left[ {{\alpha ^3} = - s} \right]\\
\Rightarrow s{\left( {1 + \alpha } \right)^3} = {r^3}\\
\Rightarrow s\left( {{\alpha ^3} + 3{\alpha ^2} + 3\alpha + 1} \right) = {r^3}\\
\Rightarrow s\left( { - s + 3\alpha \left( {\alpha + 1} \right) + 1} \right) = {r^3}\\
\Rightarrow s\left( { - s - 3r + 1} \right) = {r^3}\\
\Rightarrow {r^3} + {s^2} + 3rs - s = 0\left( {proved} \right)
\end{array}"/>