How do you simplify $\left( 8100 \right)\left( 6-6 \right)$ using order of operators?

Hint: Here in this question we have been asked to simplify the given expression $\left( 8100 \right)\left( 6-6 \right)$ using order of operators. We know that the order of operators is specified by the “BODMAS” rule which states the precedence of operators’ decreases from Bracket, of, division, multiplication, addition and subtraction.

Complete step by step solution:
Now we need to simplify the given expression $\left( 8100 \right)\left( 6-6 \right)$ using order of operators.
From the basics of algebra concepts we know that the order of operators is specified by the “BODMAS” rule which states the precedence of operators’ decreases in the given order.
(1) Bracket
(2) Of
(3) Division
(4) Multiplication
(5) Addition
(6) Subtraction.
Now we will look for the operators one by one and then perform the operations one by one. This is the process of simplification for this expression.
By further simplifying this expression we will simplify brackets first. After doing that we will have \[\Rightarrow \left( 8100 \right)\left( 6-6 \right)=8100\times \left( 6-6 \right)\]
Now we will perform the subtraction between the two numbers in the bracket. After doing that we will have $\Rightarrow 8100\times \left( 6-6 \right)=8100\times 0$ .
So the simplified expression is $\Rightarrow \left( 8100 \right)\left( 6-6 \right)=0$ .

Note: In the process of solving this question we need to take care of the calculations we perform and the concepts that we apply. Similarly we can simplify any complex expression using the “BODMAS” rule which specifies the precedence of operators. For example the simplified expression of this expression $3500\times 0$ will be $\Rightarrow 3500\times 0=0$ .