An Extension Of The Binomial Theorem

Find the coefficient of x 8 y 4 x^8y^4 in the expansion of

( ( x y ) 6 ( x + y ) 6 ) 2 \large \left((x-y)^6-(x+y)^6\right)^2


The answer is 960.

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1 solution

Chew-Seong Cheong
Feb 18, 2018

f ( x , y ) = ( ( x y ) 6 ( x + y ) 6 ) 2 = ( x y ) 12 2 ( x y ) 6 ( x + y ) 6 + ( x + y ) 12 = ( x y ) 12 2 ( x 2 y 2 ) 6 + ( x + y ) 12 = k = 0 12 ( 1 ) k ( 12 k ) x 12 k y k 2 k = 0 6 ( 1 ) k ( 6 k ) x 12 2 k y 2 k + k = 0 12 ( 12 k ) x 12 k y k \begin{aligned} f(x,y) & = \left((x-y)^6-(x+y)^6\right)^2 \\ & = (x-y)^{12}-2(x-y)^6(x+y)^6+(x+y)^{12} \\ & = (x-y)^{12}-2(x^2-y^2)^6+(x+y)^{12} \\ & = \sum_{k=0}^{12} (-1)^k \binom {12}k x^{12-k}y^k - 2 \sum_{k=0}^6 (-1)^k \binom 6k x^{12-2k}y^{2k} + \sum_{k=0}^{12} \binom {12}k x^{12-k}y^k \end{aligned}

Therefore, the coefficient of x 8 y 4 x^8y^4 is

a 8 , 4 = ( 1 ) 4 ( 12 4 ) 2 ( 1 ) 2 ( 6 2 ) + ( 12 4 ) = 2 ( 12 4 ) 2 ( 6 2 ) = 2 × 12 × 11 × 10 × 9 4 × 3 × 2 × 1 2 × 6 × 5 2 = 990 30 = 960 \begin{aligned} a_{8,4} & = (-1)^4 \binom {12}4 - 2 (-1)^2 \binom 62 + \binom {12}4 \\ & = 2 \binom {12}4 - 2\binom 62 \\ & = 2 \times \frac {12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} - 2 \times \frac {6\times 5}2 \\ & = 990 - 30 = \boxed{960} \end{aligned}

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