a³+b³+c³−3abc=(a+b+c)(a²+b²+c²−ab−bc−ca)
の公式で
(1) a→x, b→y, c→1
(2) a→a, b→-2b, c→2
と置き換える
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a³+b³+c³−3abc=(a+b+c)(a²+b²+c²−ab−bc−ca)
の公式で
(1) a→x, b→y, c→1
(2) a→a, b→-2b, c→2
と置き換える
Users viewing this question
are also looking at these questions 😉
