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Cho $a+b+c=0.$ Chứng minh rằng:$$a^3+b^3+c^3=3abc$$

score 0 · Answer 1

a^{3} + b^{3} + c^{3} = 3 a b c \Rightarrow a^{3} + b^{3} + c^{3} - 3 a b c = 0

t a c ó :

a^{3} + b^{3} + c^{3} - 3 a b c = (a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}) + c^{3} - 3 a b c - 3 a^{2} b - 3 a b^{2}

= (a + b)^{3} + c^{3} - 3 a b (a + b + c)

= (a + b + c) [(a + b)^{2} - (a + b) c + c^{2}] - 3 a b (a + b + c)

= (a + b + c) (a^{2} + 2 a b + b^{2} - a c - b c + c^{2} - 3 a b)

= (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)

vì

a + b + c = 0 \Rightarrow t h ằ n g k i a = 0 \Rightarrow (đ p c m)

a + b + c = 0 \Rightarrow t h ằ n g k i a = 0 \Rightarrow (đ p c m)

Ghost rider · Answer 2 · 10/31/2015 2:02:43 PM

đây nhá:nếu $a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+c^3-3abc=0$

$ta có:$

$a^3+b^3+c^3-3abc=(a^3+3a^2b+3ab^2+b^3)+c^3-3abc-3a^2b-3ab^2$

$=(a+b)^3+c^3-3ab(a+b+c)$

$=(a+b+c)[(a+b)^2-(a+b)c+c^2]-3ab(a+b+c)$

$=(a+b+c)(a^2+2ab+b^2-ac-bc+c^2-3ab)$

$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

vì $a+b+c=0\Rightarrow thằng kia =0\Rightarrow (đpcm)$

Trả lời 31-10-15 02:02 PM

Ghost rider
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bài này dễ mà – ๖ۣۜTQT☾♋☽ 01-11-15 09:41 PM

๖ۣۜPXM๖ۣۜMinh4212♓ · Answer 3 · 10/30/2015 9:30:07 PM

$a^3+b^3+c^3=(a+b)^3+c^3-3ab(a+b)$

$=(a+b+c)((a+b)^2-(a+b)c+c^2)-3ab(a+b)$

Vì $a+b+c=0 \Rightarrow a+b=-c$

thay vào $a^3+b^3+c^3=3abc $

Trả lời 30-10-15 09:30 PM

๖ۣۜPXM๖ۣۜMinh4212♓
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