Question

Which of the options given below is the square of the binomial $\left(8-\frac{1}{x}\right)$ ?

(i) $64-\frac{1}{x^2}$ (ii) $64+\frac{1}{x^2}$ (iii) $64-\frac{16}{x}+\frac{1}{x^2}$ (iv) $64+\frac{16}{x}+\frac{1}{x^2}$

Answer 1

The given binomial is $\left(8-\frac{1}{x}\right)$.
$$\begin{gathered} {\left(8-\frac{1}{x}\right)}^2 \\ ={\left(8\right)}^2-2\times \left(8\right)\times \left(\frac{1}{x}\right)+{\left(\frac{1}{x}\right)}^2\left[\because {\left(a-b\right)}^2=a^2-2ab+b^2\right] \\ =64-\frac{16}{x}+\frac{1}{x^2} \end{gathered}$$
Hence, the correct answer is option (iii).