xy operation (multiply) is always commutative
and the second one as well even consisting subtraction because of the square commutative as well and as we have subtraction of the very same values, both equations are 0s
So we need actually to evaluate only the 3rd one, so it's either C or E
And there is actually no need for substitution as the "-" in this equation is not a commutative operation, so it will not equal to 0
What are the odds of having the questions related to commutators
...
and the second one as well even consisting subtraction because of the square commutative as well and as we have subtraction of the very same values, both equations are 0s
So we need actually to evaluate only the 3rd one, so it's either C or E
And there is actually no need for substitution as the "-" in this equation is not a commutative operation, so it will not equal to 0
What are the odds of having the questions related to commutators
...
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