Question

Factorise copletely
a^5-a

Answer 1

Given :

Factorise

a^5-a

Solution:

Take "a" as a common

$\implies\sf a^5-a \\ \\ \implies\sf a(a^4-1)$

Applying identity to solve this question

a² - b² = (a+b)(a-b)

$\implies\sf a((a^2)^2-(1)^2) \\ \\ \implies\sf a((a^2+1)(a^2-1)) \\ \\ \implies\sf a((a^2+1)((a)^2-(1)^2) \\ \\ \implies\sf a((a^2+1)(a+1)(a-1))$

Some important identities :

• a² - b² = (a+b)(a-b)
• (a+b)² = a² + b² + 2ab
• (a-b)² = a² + b² -2ab
• a³ - b³ = (a-b)(a²+ab+b²)
• a³ + b³ = (a+b)(a²-ab+b²)
• (a+b)³ = a³ + b³ + 3ab(a+b)
• (a-b)³ = a³ - b³ - 3ab(a-b)