Giải nhanh giúp mình

H

huynhbachkhoa23

$a^3+b^3+c^3=3abc$
suy ra $a+b+c=0$
$P=(1+\frac{a}{b})(1+\frac{b}{c})(1+\frac{c}{a})$
$P=\frac{-c}{b}.\frac{-a}{c}.\frac{-b}{a}=-1$
 
C

chonhoi110

Ta có: $a^3+b^3+c^3 = 3abc$

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

$\leftrightarrow \left\{\begin{matrix}a+b+c=0\\ a^2+b^2+c^2-ab-bc-ac=0 \end{matrix}\right.$

$TH_1: a+b+c=0 \rightarrow \left\{\begin{matrix}a+b=-c\\ a+b=-c\\ b+c=-a \end{matrix}\right.$

$\rightarrow P= \dfrac{a+b}{b}.\dfrac{b+c}{c}. \dfrac{a+c}{a}= \dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}=-1$

$TH_2: a^2+b^2+c^2-ab-bc-ac=0$

$\leftrightarrow 2a^2+2b^2+2c^2-2ab-2bc-2ac=0$

$\leftrightarrow (a-b)^2+(b-c)^2+(a-c)^2=0$

$\leftrightarrow a=b=c$

$\rightarrow P=(1+1)(1+1)(1+1)=8$
 
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