Question

Cho các số nguyên a,b,c thỏa mãn: (a-b)³+(b-c)³+(c-a)³=210

Tính GT của BT: A=|a-b|+|b-c|+|c-a|

Answer 1

$(a-b{)}^3=a^3-3a^{2}b+3a{b}^2-b^3$
$(b-c{)}^3=b^3-3b^{2}c+3b{c}^2-c^3$
$(c-a{)}^3=c^3-3c^{2}a+3c{a}^2-a^3$
$=>(a-b{)}^3+(b-c{)}^3+(c-a{)}^3=-3a^{2}b+3a{b}^2-3b^{2}c+3b{c}^2-3c^{2}a+3c{a}^2=210$
$<=>-3(a^{2}b-a{b}^2+b^{2}c-b{c}^2+c^{2}a-c{a}^2)=210$
$<=>-3\left[a^2\right(b-c)+bc(b-c)-a(b^2-c^2\left)\right]=210$
$<=>-3(b-c)[a^2+bc-a(b+c\left)\right]=210$
$<=>-3(b-c)(a^2+bc-ab-ac)=210$
$<=>-3(b-c)\left[a\right(a-c)-b(a-c\left)\right]=-3(b-c)(a-c)(a-b)=210$
$<=>3(b-c)(c-a)(a-b)$
$<=>(b-c)(a-b)(c-a)=70$
$=>b-c=2,a-b=5,c-a=7$
$=>|a-b|+|b-c|+|c-a|=14$
 

Answer 2

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
Cộng ba pt, ta được
-3a^2b+3ab^2-3b^2c+3bc^2-3c^2a+3ca^2
-3(a^2b-ab^2+b^2c-bc^2+c^2a-ca^2)
-3(a^2(b-c)+bc(b-c)-a(b^2-c^2))
-3(b-c)(a^2+bc-a(b+c))
-3(b-c)(a-b)(a-c)=210
(b-c)(a-b)(a-c)=-70
(b-c)(a-b)(a-c)=2*5*(-7)
=>b-c=2, a-b=5, a-c=-7
=>|a-b|+|b-c|+|c-a|=14