If I do $3(6 + 9) = (3 \times 6) + (3 \times9)$, I get the correct answer.
But when I am doing $3 \times 6(6 + 9 -12) = (3 \times 6) + (3 \times 9) + (3 \times (-12)) + (6 \times 6) + (6 \times 9) + (6 \times(-12))$, I am getting $54$ and $27$ as the answers respectively. Please help me understand this, which rule is being applied here?
When you have
$$3*6*(6+9-12)$$
you can solve it in many different ways.
1.You first multiply the numbers in brackets by 6,then sum them up and multiply by 3.
\begin{align} &3 * 6 * (6+9-12)\\ &=3 * (6 * (6+9-12))\\ &=3 * (6 * 6+6 * 9+6*(-12))\\ &=3*(36+54-72)\\ &=3*(90-72)\\ &=3*18\\ &=54 \end{align}
2.The same as in the 1. way,but you multiply with 3 first and 6 in the end.
3 * 6*(6+9-12)=6*(3*(6+9-12))=6*(3* 6+3* 9+3*(-12))=6*(18+27-36)= =6*9=54
3.You can first sum up the numbers in brackets,then multiply by 6 and 3,which is the same as if you multiply by 18.
3* 6* (6+9-12)=18* 3=54
4.You can first multiply 6 and 3, then multiply every number in brackets by 18 and then sum up.
6* 3* (6+9-12)=18* (6+9-12)=108+162-216=54
In all four examples you get the same answer.
What was your mistake?
When you wrote
3×6(6+9−12)=(3×6)+(3×9)+(3×(−12))+(6×6)+(6×9)+(6×(−12)) the part +(6×6)+(6×9)+(6×(−12)) was wrong.You should just multiply by 6.Whenever you multiply you can use each factor only once.In your case (6+9-12) is one factor so you can multiply it only once,by3, by 6 or by 18.But you can also distribute the multiplycation over all the numbers in the sum,but only once,with one factor.
Maybe you also mistook this distribution with multiplying forms like
(a+b)*(c+d) where you actually need to multiply each number with all the others in the other brackets.In this case you could get
(3×6)+(3×9)+(3×(−12))+(6×6)+(6×9)+(6×(−12)) if the evaluation was (3+6)*(6+9-12).
If you don't understand this last paragraph just ignore it,so it doesn't confuse you further.
