20. 因式分解：
(1) $4x^2y^3 + 8x^2y^2z - 12xy^2z$；
(2) $x^3 - 4x$；
(3) $5x(x - 2y)^3 - 20y(2y - x)^3$；
(4) $2ax^2 - 12ax + 18a$.

(1)4x²y³ + 8x²y²z - 12xy²z = 4xy²(xy + 2xz - 3z).
(2)x³ - 4x = x(x² - 4)=x(x + 2)(x - 2).
(3)5x(x - 2y)³ - 20y(2y - x)³ = 5x(x - 2y)³ + 20y(x - 2y)³ = 5(x - 2y)³(x + 4y).
(4)2ax² - 12ax + 18a = 2a(x² - 6x + 9)=2a(x - 3)².

21. 如图,已知 $\triangle ABC$ 为等边三角形,点 $D$,$E$ 分别在边 $BC$,$AC$ 上,且 $AE = CD$,$AD$ 与 $BE$ 相交于点 $F$.
(1) 求证：$\triangle ABE ≌ \triangle CAD$；
(2) 求 $∠BFD$ 的度数.

(1)证明:因为△ABC为等边三角形.所以∠BAE = ∠C = 60°,AB = CA.在△ABE和△CAD中,$\left\{\begin{array}{l} AE = CD,\\ ∠BAE = ∠C,\\ AB = CA,\end{array}\right. $所以△ABE≌△CAD(SAS).
(2)解:由
(1),知△ABE≌△CAD,所以∠ABE = ∠CAD,所以∠BFD = ∠ABE + ∠BAD = ∠CAD + ∠BAD = ∠BAC = 60°.