Question

Evaluate:
$(a-\frac{1}{a})(a+\frac{1}{a})(a^2+\frac{1}{a^2})(a^4+\frac{1}{a^4})(a^8+\frac{1}{a^8})$

Answer 1

As we know that $(a+b)(a-b)=a^2-b^2$

$(a-\frac{1}{a})(a+\frac{1}{a})(a^2+\frac{1}{a^2})(a^4+\frac{1}{a^4})(a^8+\frac{1}{a^8})=(a^2-\frac{1}{a^2})(a^2+\frac{1}{a^2})(a^4+\frac{1}{a^4})(a^8+\frac{1}{a^8})$

$=(a^4-\frac{1}{a^4})(a^4+\frac{1}{a^4})(a^8+\frac{1}{a^8})=(a^8-\frac{1}{a^8})(a^8+\frac{1}{a^8})=(a^{16}-\frac{1}{a^{16}})$