Question

1,Biết a + b + c = 0. Chứng minh a3 + b3 + c3 = 3abc.

Answer 1

Ta có: a + b + c = 0

⇒ a + b = -c ⇒ (a + b)3 = (-c)3

⇒ a3 + b3 + 3ab(a + b) = -c3 ⇒ a3 + b3 + 3ab(-c) + c3 = 0

⇒ a3 + b3 + c3 = 3abc

Answer 2

a+b+c=0

\(\Rightarrow\)(a+b+c)\(^3\)=0

\(\Rightarrow\)a\(^3\)+b\(^3\)+c\(^3\)+3a\(^2\)b+3ab\(^2\)+3b\(^2\)c+3bc\(^2\)+3a\(^2\)c+3ac\(^2\)+6abc=0

\(\Rightarrow\)a\(^3\)+b\(^3\)+c\(^3\)+(3a\(^2\)b+3ab\(^2\)+3abc)+(3b\(^2\)c+3bc\(^2\)+3abc)+(3a\(^2\)c+3ac\(^2\)+3abc)-3abc=0

=>a\(^3\)+b\(^3\)+c\(^3\)+3ab(a+b+c)+3bc(a+b+c)+3ac(a+b+c)=3abc

Do a+b+c=0

=>a\(^3\)+b\(^3\)+c\(^3\)=3abc(ĐPCM)

Answer 3

Ta có

a³ + b³ + c³= 3abc

⇒(a+b)³ − 3ab(a+b) + c³ − 3abc = 0

⇒(a + b + c)[(a + b)² − (a + b)c + c²] − 3ab(b + c + a)=0

⇒(a + b + c)(a² + b² + c² − ab − bc − ca)=0

Mà đẳng thức (a+b+c)(a²+b²+c² - ab - bc ca ) = 0 đúng vì a+b+c = 0

=> a³+b³+c3=3abc

Answer 4

Theo đề ta có:
\(a+b+c=0\) \(\Rightarrow c=-(a+b) (1)\)
Thay \((1)\) vào \( a^3+b^3+c^3\) ta có:
\(a^3+b^3+[-(a+b)]^3=3ab[-(a+b)]\\ \Leftrightarrow a^3+b^3-(a+b)=-3ab(a+b)\\ \Leftrightarrow a^3+ b^3- a^3 -3a^2b- 3ab^2- b^3= -3a^2b- 3ab^2\\ \Leftrightarrow 0= 0 \)
Vậy ta có đpcm.

Answer 5

Mơn mấy bn nhaaa !