উদাহরণ 1. [tex]2{a^3} + 6a - 3 = 0[/tex] হলে দেখাও যে [tex]a = {2^{\frac{1}{3}}} - {2^{ - \frac{1}{3}}}[/tex] হয়। [H.S. '85]
সমাধান :
[tex]\begin{array}{l}
2{a^3} + 6a - 3\\
= 2{\left( {{2^{\frac{1}{3}}} - {2^{ - \frac{1}{3}}}} \right)^3} + 6\left( {{2^{\frac{1}{3}}} - {2^{ - \frac{1}{3}}}} \right) - 3\\
= 2\{ {\left( {{2^{\frac{1}{3}}}} \right)^3} - 3{\left( {{2^{\frac{1}{3}}}} \right)^2}\left( {{2^{ - \frac{1}{3}}}} \right) + 3\left( {{2^{\frac{1}{3}}}} \right){\left( {{2^{ - \frac{1}{3}}}} \right)^3} - {\left( {{2^{ - \frac{1}{3}}}} \right)^3}\} + 6\left( {{2^{\frac{1}{3}}} - {2^{ - \frac{1}{3}}}} \right) - 3\\
= 2\left\{ {{2^{\frac{1}{3} \times 3}} - 3 \times {2^{\frac{2}{3} - \frac{1}{3}}} + 3 \times {2^{\frac{1}{3} - \frac{2}{3}}} - {2^{ - \frac{1}{3} \times 3}}} \right\} + 6\left( {{2^{\frac{1}{3}}} - {2^{ - \frac{1}{3}}}} \right) - 3\\
= 2\left\{ {2 - 3 \times {2^{\frac{1}{3}}} + 3 \times {2^{ - \frac{1}{3}}} - {2^{ - 1}}} \right\} + 6\left( {{2^{\frac{1}{3}}} - {2^{ - \frac{1}{3}}}} \right) - 3\\
= 2\left( {2 - \frac{1}{2}} \right) - 6\left( {{2^{\frac{1}{3}}} - {2^{ - \frac{1}{3}}}} \right) + 6\left( {{2^{\frac{1}{3}}} - {2^{ - \frac{1}{3}}}} \right) - 3\\
= 2 \times \frac{3}{2} - 3\\
= 3 - 3\\
= 0
\end{array}[/tex]
