Câu hỏi của Nguyễn Trọng Phúc

Question

CM: $a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)$

Hoang Hung Quan · Accepted Answer

Giải:

Ta có: $VT=a^3+b^3+c^3-3abc$

$=\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc$

$=\left(a+b+c\right)^3\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b\right)-3abc$

$=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2\right)-3ab\left(a+b+c\right)$

$=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2-3ab\right)$

$=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)$

$=VP$ (Đpcm)Câu trả lời: Giải:

Ta có: $VT=a^3+b^3+c^3-3abc$

$=\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc$

$=\left(a+b+c\right)^3\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b\right)-3abc$

$=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2\right)-3ab\left(a+b+c\right)$

$=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2-3ab\right)$

$=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)$

$=VP$ (Đpcm)

Đức Hiếu · Answer

Ta có:

$VP=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)$

$=a^3+ab^2+ac^2-a^2b-abc-a^2c+a^2b+b^3+bc^2-ab^2-b^2c-abc+a^2c+b^2c+c^3-abc-bc^2-ac^2$

$=a^3+b^3+c^3-3abc=VT$

$\rightarrow$ đpcm

Chúc bạn học tốt!!!Câu trả lời: Ta có:

$VP=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)$

$=a^3+ab^2+ac^2-a^2b-abc-a^2c+a^2b+b^3+bc^2-ab^2-b^2c-abc+a^2c+b^2c+c^3-abc-bc^2-ac^2$

$=a^3+b^3+c^3-3abc=VT$

$\rightarrow$ đpcm

Chúc bạn học tốt!!!