Cho các số nguyên a,b,c thỏa mãn (a-b)^3+(b-c)^3+(c-a)^3=210. Tính A=/a-b/+/b-c/+/c-a/

M

minhmai2002

Ta có:
$(a-b)^3=a^3-3a^2b+3ab^2-b^3$
$(b-c)^3=b^3-3b^2c+3bc^2-c^3$
$(c-a)^3=c^3-3c^2a+3ca^2-a^3$
$=> (a-b)^3+(b-c)^3+(c-a)^3=-3a^2b+3ab^2-3b^2c+3bc^2-3c^2a+3ca^2=210$
$<=> -3(a^2b-ab^2+b^2c-bc^2+c^2a-ca^2)=210$
$<=> -3[a^2(b-c)+bc(b-c)-a(b^2-c^2)]=210$
$<=> -3(b-c)[a^2+bc-a(b+c)]=210$
$<=> -3(b-c)(a^2+bc-ab-ac)=210$
$<=> -3(b-c)[a(a-c)-b(a-c)]=-3(b-c)(a-c)(a-b)=210$
$<=> 3(b-c)(c-a)(a-b)$
$<=> (b-c)(a-b)(c-a)=70$
$=>b-c=2, a-b=5, c-a=7$
$=>|a-b|+|b-c|+|c-a|=14$
 
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