Answer
$$(2+i)(2-i)(4+3i)=20+15i$$
Work Step by Step
$$A=(2+i)(2-i)(4+3i)$$
For $(2+i)(2-i)$, we can still use the formula $$(a+b)(a-b)=a^2-b^2$$ which means $$(2+i)(2-i)$$ $$=4-i^2$$ $$=4-(-1)$$ $$=5$$
Now we can apply the result just found to $A$ $$A=5(4+3i)$$ $$A=20+15i$$