Câu hỏi của Kiệt Nguyễn

Question

$CMR:a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)$

Phạm Thị Thùy Linh · Accepted Answer

$a^3+b^3+c^3-3abc$$=\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-\left(3a^2b+3ab^2+3abc\right)$$=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)$$=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)$$=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2-3ab\right)$$=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)$$\left(đpcm\right)$Câu trả lời: $a^3+b^3+c^3-3abc$$=\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-\left(3a^2b+3ab^2+3abc\right)$$=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)$$=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)$$=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2-3ab\right)$$=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)$$\left(đpcm\right)$

Kiên-Messi-8A-Boy2k6 · Answer

$a^3+b^3+c^3-3abc$$=\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc$$=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)$$=\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2\right)-3ab\left(a+b+c\right)$$=\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2-3ab\right)$$=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2-3ab\right)$$=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)$Câu trả lời: $a^3+b^3+c^3-3abc$$=\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc$$=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)$$=\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2\right)-3ab\left(a+b+c\right)$$=\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2-3ab\right)$$=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2-3ab\right)$$=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)$

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