①(2y-z)²[2y(z+2y)+z²]²=②（-x-y)²(x²-xy+y²)²=

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①(2y-z)²[2y(z+2y)+z²]²=②（-x-y)²(x²-xy+y²)²=①(2y-z)²[2y(z+2y)

①(2y-z)²[2y(z+2y)+z²]²=②（-x-y)²(x²-xy+y²)²=
①(2y-z)²[2y(z+2y)+z²]²=
②（-x-y)²(x²-xy+y²)²=

①(2y-z)²[2y(z+2y)+z²]²=②（-x-y)²(x²-xy+y²)²=
解①
原式={(2y-z)×[2y(z+2y)+z²]}²
={(2y-z)[(2y)²+2yz+z²]}²
=[(2y)³-z³]²
=(8y³-z³)²
=64y^6-16y³z³+z^6

解②
原式=(x+y)²(x²-xy+y²)²
=[(x+y)(x²-xy+y²)]²
=(x³+y³)²
=x^6+2x³y³+y^6