Question

factorize a¹⁶-1 .fast​

Answer 1

Answer:

a¹⁶-1 = $(a^{8} + 1) (a^{4} + 1) (a^{2} + 1) (a + 1) (a - 1)$

Step-by-step explanation:

here, we solve it with a² - b² = (a+b) (a-b) formula.

so, a¹⁶-1

= $(a^{8} + 1) (a^{8} - 1)$

= $(a^{8} + 1)$ $(a^{4} +1) (a^{4} - 1)$          (∵ 1² = 1)

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

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

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

∴ a¹⁶-1 = $(a^{8} + 1) (a^{4} + 1) (a^{2} + 1) (a + 1) (a - 1)$