How to simplify $A(\overline BC+B)$

Question

How do I go from $A(\overline BC+B)$ to $A(B+C)$? What definition should I use to get the final answer?

Would like an explanation and proof so I can learn rather than just memorise.

Answer 1

$\displaystyle\bar BC+B=\bar BC+B\cdot1=\bar BC+B(1+C)$ as $1+C=1$

$\displaystyle\implies \bar BC+B=C(B+\bar B)+B=C\cdot1+B=C+B$

Answer 2

use the distributive law $$(B'C+B)=(B+B').(B+C)$$ also $$B+B'=1$$