Câu hỏi của ๛•Ŧ๏кเϮ๏ų~ Ɱʉїċɦїɾøʉツ

Question

Cho a + b + c = 0 . Chứng minh rằng : a3 + b3 + c3 = 3abc

I - Vy Nguyễn · Accepted Answer

Ta có : $a+b+c=0$$\Rightarrow\hept{\begin{cases}a+b=-c\b+c=-a\a+c=-b\end{cases}}$ ( 1 )Ta có : $a+b+c=0$$\Rightarrow\left(a+b+c\right)^3=0$$\Rightarrow\left[\left(a+b\right)+c\right]^3=0$$\Rightarrow\left(a+b\right)^3+c^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2=0$$\Rightarrow\left(a+b\right)^3+c^3+3\left(a+b\right)\left[\left(a+b\right)c+c^2\right]=0$$\Rightarrow\left(a+b\right)^3+c^3+3\left(a+b\right)c\left(a+b+c\right)=0$$\Rightarrow a^3+b^3+3a^2b+3ab^2+c^3+3\left(a+b\right)c\left(a+b+c\right)=0$$\Rightarrow a^3+b^3+3ab\left(a+b\right)+c^3+3\left(a+b\right)c\left(a+b+c\right)=0$$\Rightarrow a^3+b^3+c^3+3\left(a+b\right)\left[ab+c\left(a+b+c\right)\right]=0$$\Rightarrow a^3+b^3+c^3+3\left(a+b\right)\left(ab+ca+cb+c^2\right)=0$$\Rightarrow a^3+b^3+c^3+3\left(a+b\right)\left[\left(ab+ca\right)+\left(cb+c^2\right)\right]=0$$\Rightarrow a^3+b^3+c^3+3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]=0$$\Rightarrow a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)=0$ ( 2 ) Thay ( 1 ) vào ( 2 ) ta được :  $\Rightarrow a^3+b^3+c^3+3.\left(-c\right).\left(-a\right).\left(-b\right)=0$$\Rightarrow a^3+b^3+c^3-3abc=0$$\Rightarrow a^3+b^3+c^3=3abc$Câu trả lời: Ta có : $a+b+c=0$$\Rightarrow\hept{\begin{cases}a+b=-c\b+c=-a\a+c=-b\end{cases}}$ ( 1 )Ta có : $a+b+c=0$$\Rightarrow\left(a+b+c\right)^3=0$$\Rightarrow\left[\left(a+b\right)+c\right]^3=0$$\Rightarrow\left(a+b\right)^3+c^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2=0$$\Rightarrow\left(a+b\right)^3+c^3+3\left(a+b\right)\left[\left(a+b\right)c+c^2\right]=0$$\Rightarrow\left(a+b\right)^3+c^3+3\left(a+b\right)c\left(a+b+c\right)=0$$\Rightarrow a^3+b^3+3a^2b+3ab^2+c^3+3\left(a+b\right)c\left(a+b+c\right)=0$$\Rightarrow a^3+b^3+3ab\left(a+b\right)+c^3+3\left(a+b\right)c\left(a+b+c\right)=0$$\Rightarrow a^3+b^3+c^3+3\left(a+b\right)\left[ab+c\left(a+b+c\right)\right]=0$$\Rightarrow a^3+b^3+c^3+3\left(a+b\right)\left(ab+ca+cb+c^2\right)=0$$\Rightarrow a^3+b^3+c^3+3\left(a+b\right)\left[\left(ab+ca\right)+\left(cb+c^2\right)\right]=0$$\Rightarrow a^3+b^3+c^3+3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]=0$$\Rightarrow a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)=0$ ( 2 ) Thay ( 1 ) vào ( 2 ) ta được :  $\Rightarrow a^3+b^3+c^3+3.\left(-c\right).\left(-a\right).\left(-b\right)=0$$\Rightarrow a^3+b^3+c^3-3abc=0$$\Rightarrow a^3+b^3+c^3=3abc$

☆MĭηɦღAηɦ❄ · Answer

$a^3 + b^3 + c^3 = (a+b)(a^2-ab+b^2) + 3ab(a+b) + c^3 - 3ab(a+b)$$= (a+b)^3 + c^3 - 3ab(a+b)$$= (a+b+c)(a^2 + 2ab + b^2 + ac + bc + c^2) - 3ab(a+b)
$$= 0 - 3ab(a+b)$Từ $a+b+c = 0 => a+b = -c$Thay vào ta được : $-3ab(a+b) = -3ab(-c) = 3abc$Lẹ hơn xíu ~Câu trả lời: $a^3 + b^3 + c^3 = (a+b)(a^2-ab+b^2) + 3ab(a+b) + c^3 - 3ab(a+b)$$= (a+b)^3 + c^3 - 3ab(a+b)$$= (a+b+c)(a^2 + 2ab + b^2 + ac + bc + c^2) - 3ab(a+b)
$$= 0 - 3ab(a+b)$Từ $a+b+c = 0 => a+b = -c$Thay vào ta được : $-3ab(a+b) = -3ab(-c) = 3abc$Lẹ hơn xíu ~

Chúng tôi đề xuất

Tài nguyên

Ứng dụng mobile

Bảng xếp hạng

Các khóa học có thể bạn quan tâm

Yêu cầu VIP