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(x+m)的四次方-(x-m)的四次方
=[(x+m)²]²-[(x-m)²]²
=[(x+m)²+(x-m)²][(x+m)²-(x-m)²]
=[(x²+2mx+m²)+(x²-2mx+m²)][(x²+2mx+m²)-(x²-2mx+m²)]
=(2x²+2m²)(4mx)
=8mx(x²+m²)

原式=[(x + m)~2 +（x - m)~2]*[(x + m)~2 - (x - m)~2]
=8mx(x~2 + m~2)

注：x~2为x的平方；（x - m)~2为（x - m)~2的平方

2项式定理 2x^2 m^2 + 2 x m^3

(x+m)^4 - (x-m)^4
=[(x+m)^2 + (x-m)^2]*[(x+m)^2 - (x-m)^2]
=[(x^2+2mx+m^2)+(x^2-2mx+m^2)]*[(x^2+2mx+m^2)-(x^2-2mx+m^2)]
=2(x^2+m^2)*4mx
=8mx(x^2+m^2)

=[(x+m)²]²-[(x-m)²]²
=[(x+m)²+(x-m)²][(x+m)²-(x-m)²]
=[(x²+2mx+m²)+(x²-2mx+m²)][(x²+2mx+m²)-(x²-2mx+m²)]
=(2x²+2m²)(4mx)
=8mx(x²+m²)