MHB How Can You Expand a High-Power Binomial Expression Using a Shortcut Technique?

[tiffanyfm]

expand binomial (2x^2-2y^2)^9

 

[MarkFL]

expand binomial (2x^2-2y^2)^9

I think I would write:

$$S=\left(2x^2-2y^2\right)^9=2^9\left(x^2-y^2\right)^9=512\left(x^2+(-y^2)\right)^9$$

Now, the binomial theorem tells us we may write:

$$S=512\sum_{k=0}^{9}\left({9 \choose k}(x^2)^{9-k}(-y^2)^k\right)$$

I was taught a shortcut for binomial expansions when I was a student, which I will demonstrate as follows:

Suppose we wish to expand $$(x+y)^8$$...so we set up our first term with a coefficient of 1, the first term in the binomial being expanded gets an 8 as an exponent, while the second term gets a 0:

$$(x+y)^8=1\cdot x^8y_0+\cdots$$

Now, to compute the coefficient of the second term, we use the first term...we take the exponent on the first term of the binomial being expanded, multiply it by the coefficient on the first term, and divide by the number of the term, which for the first term is 1...and for the two terms we take the exponent on the first down by one and we bump up the exponent on the second...hence:

$$(x+y)^8=1\cdot x^8y^0+\frac{8\cdot1}{1}x^7y^1+\cdots$$

Let's clean it up a bit...

$$(x+y)^8=x^8+8x^7y+\cdots$$

Now do the same thing to get the third term:

$$(x+y)^8=x^8+8x^7y+\frac{7\cdot8}{2}x^6y^2+\cdots$$

$$(x+y)^8=x^8+8x^7y+28x^6y^2+\cdots$$

And so forth, until we get to the 5th term, at which point we know by symmetry the binomial coefficients will be reflected across that term, until we have:

$$(x+y)^8=x^8+8x^7y+28x^6y^2+56x^5y^3+70x^4y^4+56x^3y^5+28x^2y^6+8xy^7+y^8$$

Can you apply this technique to the problem at hand?