সূচক সংক্রান্ত অংকের সমাধান (Solution of Indices )

উদাহরণ 1. $2{a^3} + 6a - 3 = 0$ হলে দেখাও যে $a = {2^{\frac{1}{3}}} - {2^{ - \frac{1}{3}}}$ হয়। [H.S. '85]

সমাধান : 

$\begin{array}{l} 2{a^3} + 6a - 3\\  = 2{\left( {{2^{\frac{1}{3}}} - {2^{ - \frac{1}{3}}}} \right)^3} + 6\left( {{2^{\frac{1}{3}}} - {2^{ - \frac{1}{3}}}} \right) - 3\\  = 2\{ {\left( {{2^{\frac{1}{3}}}} \right)^3} - 3{\left( {{2^{\frac{1}{3}}}} \right)^2}\left( {{2^{ - \frac{1}{3}}}} \right) + 3\left( {{2^{\frac{1}{3}}}} \right){\left( {{2^{ - \frac{1}{3}}}} \right)^3} - {\left( {{2^{ - \frac{1}{3}}}} \right)^3}\}  + 6\left( {{2^{\frac{1}{3}}} - {2^{ - \frac{1}{3}}}} \right) - 3\\  = 2\left\{ {{2^{\frac{1}{3} \times 3}} - 3 \times {2^{\frac{2}{3} - \frac{1}{3}}} + 3 \times {2^{\frac{1}{3} - \frac{2}{3}}} - {2^{ - \frac{1}{3} \times 3}}} \right\} + 6\left( {{2^{\frac{1}{3}}} - {2^{ - \frac{1}{3}}}} \right) - 3\\  = 2\left\{ {2 - 3 \times {2^{\frac{1}{3}}} + 3 \times {2^{ - \frac{1}{3}}} - {2^{ - 1}}} \right\} + 6\left( {{2^{\frac{1}{3}}} - {2^{ - \frac{1}{3}}}} \right) - 3\\  = 2\left( {2 - \frac{1}{2}} \right) - 6\left( {{2^{\frac{1}{3}}} - {2^{ - \frac{1}{3}}}} \right) + 6\left( {{2^{\frac{1}{3}}} - {2^{ - \frac{1}{3}}}} \right) - 3\\  = 2 \times \frac{3}{2} - 3\\  = 3 - 3\\  = 0 \end{array}$