This problem uses the binomial expansion formula: (x+y)ⁿ= _k=0Zⁿ _nC_kx^k y^n-k.But lets try to understand where this formula comes from by doing the problem without it to start of with.We have the expression:(x²-1/x)⁹=(x²-x^-1)⁹=(x²-x^-1)(x²-x^-1)(x²-x^-1)(x²-x^-1)(x²-x^-1)(x²-x^-1)(x²-x^-1)(x²-x^-1)(x²-x^-1)If we wanted to compute all the terms in this expression, we would have to add up each different combination of multiplying one number from each of the nine brackets. However we are only looking for the constant term, aka where the x's cancel out. Since we are multiplying x²s and x^-1s, by adding exponents we can find that to be when we choose 3 x²s and 6 x^-1s. But there are many ways to choose 3 x²s and 6 x^-1s from the 9 brackets, so we need to add them up to get the final result.Fortunately, there is an easy way to do this, using Pascal's triangle. If we make a Pascal's triangle down to the 9th row and we take the 3rd number on that row, we will find out coefficient. We can also calculate this value quicker by using the formula n!/k!n! where n is the power (9) and k is the number of times we choose one of the terms in the bracket.So now lets see how that brings us back to our binomial expansion formula.We can now see that x^ky^n-k just refers to the value we get if we choose k x's._nC_k is the coefficient that we found from Pascal's Triangle to account for all the possible ways to get k x's.And finally the formula is a sum of this equation for all possible k's so that we find all the terms of the expression!

Differentiate x^x

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