Question

The factorisation of $a^4-1$ is equal to .

• A. $(a+1)(a-1)(a^2+1)$
• B. $(a-1)(a-1)(a^2+1)$
• C. $(a+1)(a+1)(a^2+1)$
• D. $(a+1)(a-1)(a^2-1)$

Answer 1

The correct option is A $(a+1)(a-1)(a^2+1)$

The given expression is $a^4-1$
It can be written as
$(a^2{)}^2-1$
= $(a^2{)}^2-1^2$
= $(a^2-1)(a^2+1)$ [Using $x^2-y^2=(x+y)(x-y)$]
= $(a-1)(a+1)(a^2+1)$ [Using $x^2-y^2=(x+y)(x-y)$]
= $(a-1)(a+1)(a^2+1)$