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pinkylun

Bài 1:

a) $a^3+b^3+c^3=3abc$

$=>(a+b)^3-3ab(a+b)+c^3-3abc=0$

$=>(a+b+c)^3-3c(a+b)(a+b+c)-3ab(a+b+c)=0$

$=>(a+b+c)[(a+b+c)^2-3ac-3bc-3ab]=0$

$=>(a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0$

$=>a+b+c=0$ :)

hoăc $a^2+b^2+c^2-ab-bc-ac=0$

$<=>2a^2+2b^2+2c^2-2ab-2bc-2ac=0$

$<=>(a-b)^2+(b-c)^2+(c-a)^2=0$ hay $a=b=c$ :))

từ :):)) =>đpcm
 
V

vanmanh2001

Bài b
Đặt $b-c = m , c-a = n , a-b = p$
Ta có $m+n+p = 0$
$\Rightarrow (m+n+p)^3 = 0$
$m^3 + n^3 + p^3 + 3(m+n)(n+p)(p+m) = 0$
$\Rightarrow m^3+n^3+p^3 = -3(m+n)(n+p)(p+m) = -3mnp$ (vì m+n+p = 0) (đpcm)
Bài c
$\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = 0$
$\Rightarrow \dfrac{1}{a^3} + \dfrac{1}{b^3} + \dfrac{1}{c^3} = \dfrac{3}{abc}$
$P = \dfrac{abc}{a^3} + \dfrac{abc}{b^3} + \dfrac{abc}{c^3}$
$= abc( \dfrac{1}{a^3} + \dfrac{1}{b^3} + \dfrac{1}{c^3}) = \dfrac{abc}{3abc} = \dfrac{1}{3}$
d)
$a^3 + b^3 + c^3 + d^3$
$= (a+b)^3 - 3ab(a+b) + (c+d)^3 - 3cd(c+d)$
$= (a+b)^3 - 3ab(a+b) - (a+b)^3 + 3cd(a+b)$
$= 3(a+b)(cd-ab)$ (đpcm)
 
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