দেওয়া আছে, $$ 4a^2+\frac{1}{a^2} = 2 $$ আমরা জানি, $$ \begin{aligned} 8a^3+\frac{1}{a^3} &= (2a)^3 + \left(\frac{1}{a}\right)^3 \\ &= \left(2a + \frac{1}{a}\right) \left\{(2a)^2 - 2a \cdot \frac{1}{a} + \left(\frac{1}{a}\right)^2\right\} \\ &= \left(2a + \frac{1}{a}\right) \left(4a^2 - 2 + \frac{1}{a^2}\right) \\ &= \left(2a + \frac{1}{a}\right) \left\{\left(4a^2 + \frac{1}{a^2}\right) - 2\right\} \end{aligned} $$ এখন, $4a^2+\frac{1}{a^2} = 2$ মান বসিয়ে পাই, $$ \begin{aligned} 8a^3+\frac{1}{a^3} &= \left(2a + \frac{1}{a}\right) (2 - 2) \\ &= \left(2a + \frac{1}{a}\right) \times 0 \\ &= 0 \end{aligned} $$ সুতরাং সঠিক উত্তর **A**।