Question

Find the product using a suitable identity: $\left(A+3\right)\left(A-3\right)\left(A^2+9\right)$

Answer 1

Simplify using the algebraic quantities:

The given expression is $\left(A+3\right)\left(A-3\right)\left(A^2+9\right)$.

It can be simplified as follows,

$\left(A+3\right)\left(A-3\right)\left(A^2+9\right)=\left[\left(A+3\right)\left(A-3\right)\right]\left(A^2+9\right)$

$\Rightarrow \left(A+3\right)\left(A-3\right)\left(A^2+9\right)=\left[{\left(A\right)}^2-{\left(3\right)}^2\right]\left(A^2+9\right)$ $[\because (a+b)(a-b)=a^2-b^2]$

$\Rightarrow \left(A+3\right)\left(A-3\right)\left(A^2+9\right)=\left(A^2-9\right)\left(A^2+9\right)$

$\Rightarrow \left(A+3\right)\left(A-3\right)\left(A^2+9\right)={\left(A^2\right)}^2-{\left(9\right)}^2$ $[\because (a-b)(a+b)=a^2-b^2]$

$\Rightarrow \left(A+3\right)\left(A-3\right)\left(A^2+9\right)=A^{2\times 2}-81$ $\left[\because {\left(a^m\right)}^n=a^{m\times n}\right]$

$\Rightarrow \left(A+3\right)\left(A-3\right)\left(A^2+9\right)=A^4-81$

Hence, the product of $\left(A+3\right)\left(A-3\right)\left(A^2+9\right)$ is $A^4-81$.