This question requires the student to think critically. It is more of a challenge problem than a typical Grade 8 math problem.
The student needs to be able to see the “trick” or “key” that enables them to do this problem.
It is obvious that if the student were to multiply all the terms out, it would take too long and is incredibly difficult.
The student should realize that starting with the second numerator and denominator, each adjacent pair is equal. Thus, the student can cancel out adjacent numerators and denominators.
The resulting expression is: