Question

Factorise the given polynomial:

$a^4-81$

Answer 1

Factorisation by using algebraic identities:

Consider the given polynomial $a^4-81$,

$$\begin{gathered} \because a^4-81={\left(a^2\right)}^2-{\left(9\right)}^2 \\ =\left(a^2-9\right)\left(a^2+9\right)[\because (x^2-y^2)=\left(x-y\right)\left(x+y\right)] \\ =\left(a^2-{\left(3\right)}^2\right)\left(a^2+9\right) \\ =\left(a-3\right)\left(a+3\right)\left(a^2+9\right)[\mathrm{Again}\mathrm{by}\mathrm{using}(x^2-y^2)=\left(x-y\right)\left(x+y\right)] \end{gathered}$$

Hence, $a^4-81$$=$$\left(a-3\right)\left(a+3\right)\left(a^2+9\right)$.