Question

$$\begin{gathered} 78.Theexpression\left(x-\frac{1}{x}\right)\left(x+\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}\right)\left(x^4+\frac{1}{x^4}\right)isequalto \\ \left(A\right)x^6-\frac{1}{x^6}\left(B\right)x^8-\frac{1}{x^8} \\ \left(C\right)x^7-\frac{1}{x^4}\left(D\right)x^{16}-\frac{1}{x^{16}} \end{gathered}$$

Answer 1

Dear student

$$\begin{gathered} \left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)\left({\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}\right)\left({\mathrm{x}}^4+\frac{1}{{\mathrm{x}}^4}\right) \\ =\left({\mathrm{x}}^2-\frac{1}{{\mathrm{x}}^2}\right)\left({\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}\right)\left({\mathrm{x}}^4+\frac{1}{{\mathrm{x}}^4}\right)\left[\because {\mathrm{a}}^2-{\mathrm{b}}^2=(\mathrm{a}-\mathrm{b})(\mathrm{a}+\mathrm{b})\right] \\ =\left({\mathrm{x}}^4-\frac{1}{{\mathrm{x}}^4}\right)\left({\mathrm{x}}^4+\frac{1}{{\mathrm{x}}^4}\right) \\ =\left({\mathrm{x}}^8-\frac{1}{{\mathrm{x}}^8}\right) \end{gathered}$$

Regards