(x−y)³+(y+z)³−(x−z)³(x−y)³+(y+z)³−(x−z)³
⇔x³−3x²y+3xy²−y³+y³+3y²z+3yz²+z³−(x³−3x²z+3xz²−z³)
⇔x³−3x²y+3xy²−y³+y³+3y²z+3yz²+z³−(x³−3x²z+3xz²−z³)
⇔x³−3x²y+3xy²+3y²z+3yz²+z³+3x²z−3xz²+z³
⇔x³−3x²y+3xy²+3y²z+3yz²+z³+3x²z−3xz²+z³
⇒−3x²y+3xy²+3y²z+3yz²+2z³+3y²z+3x²z−3xz²
⇒−3x²y+3xy²+3y²z+3yz²+2z³+3y²z+3x²z−3xz²
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