a+b/a2-b2·(a3+b3/a+b + 2ab)
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a+b/a2-b2·(a3+b3/a+b + 2ab)
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![a+b/a2-b2·(a3+b3/a+b + 2ab)](/uploads/image/z/8343684-36-4.jpg?t=a%2Bb%2Fa2-b2%C2%B7%28a3%2Bb3%2Fa%2Bb+%2B+2ab%29)
=(a+b)(a-b)/(a+b)×[(a+b)(a²-ab+b²)/(a+b)+2ab]
=(a-b)×(a²+ab+b²)
=a³-b³
=(a-b)×(a²+ab+b²)
=a³-b³
a+b/a2-b2·(a3+b3/a+b + 2ab)
[1/(a-b)-(a+b)/(a2+ab+b2)+ab/(b3-a3)]×(a3-b3)
证明:a3+b3=(a+b)(a2-ab+b2)
:(a-b)(a2 ab b2)=a3-b3 怎么解
a3+b3=(a+b)(a2-ab+b2) ?
a2-b2=(a+b)(a-b) a3+b3=(a+b)(a2-ab+b2) a3-b3=(a-b)(a2+ab+b2)
已知ab∈R+,并且a≠b,求证a3/b2+b3/a2>a+b
是不是这样的a6-b6=(a3+b3)(a3-b3)=(a+b)((a2-ab+b2)(a-b)(a2+ab+b2)=(
求(a+b)(a2+b2)(a3+b3)≥8a3b3
求证(a+b)(a2+b2)(a3+b3)>=8a3b3
如何推导a3+b3=(a+b)(a2-ab+b2)
已知实数a,b≥0,求证:a3+b3≥√ab(a2+b2)