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Theo bài ra, ta có: a+b+c
Suy ra: 3(a+b+c)-3abc=0
Suy ra: -3abc=0
Tương đương: -3*(b+c)*(a+c)*(a+b)=0
Tương đương: -3* a^2+b^2+c^2=0
Tương đương: -3*0=0
Suy ra: nếu a+b+c=0 thì a3+b3+c3-3abc=0(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A = a + b, B = c. Áp dụng hằng đẳng thức ( A + B ) 3 để biến đổi vế trái.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Áp dụng nhiều lần công thức \(\left(x+y\right)^3=x^3-y^3+3xy\left(x+y\right)\), ta có:
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)
\(=a^3+b^3+3ab\left(a+b\right)+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=3\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(Đpcm\right)\)
b) Ta có:
\(a^3+b^3+c^3-3abc\)
\(=a^3+3ab\left(a+b\right)+b^2+c^3-3abc-3ab\left(a+b\right)\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc-ab\right)\)
Mình nghĩ bằng thế này mới đúng, bạn chắc ghi sai đề rồi
a) Ta có: (a + b + c)3 - a3 - b3 - c3 = [ (a + b + c)3 - a3 ] - ( b3 + c3)
= (a + b + c - a) ( a2 + b2 + c2 + 2ab + 2bc + 2ac + a2 + ab + ac + a2) - (b + c) ( b2 - bc + c3)
= (b + c) ( 3a2 + b2 + c2 + 3ab + 2bc + 3ac) - (b + c) ( b2 - bc + c3)
= ( b + c) ( 3a2 + b2 + c2 + 3ab + 2bc + 3ac - b2 + bc - c3)
= ( b + c) ( 3a2 + 3ab + 3bc + 3ac)
= 3 (b + c) [a (a + b) + c (a + b)]
= 3 (b + c) (a + b) (a + c) (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
a )
`VP= (a+b)^3-3ab(a+b)`
`=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2`
`=a^3+b^3 =VT (đpcm)`
b)
b) Ta có
`VT=a3+b3+c3−3abc`
`=(a+b)3−3ab(a+b)+c3−3abc`
`=[(a+b)3+c3]−3ab(a+b+c)`
`=(a+b+c)[(a+b)2+c2−c(a+b)]−3ab(a+b+c)`
`=(a+b+c)(a2+b2+2ab+c2−ac−bc−3ab)`
`=(a+b+c)(a2+b2+c2−ab−bc−ca)=VP`
a) Ta có:
`VP= (a+b)^3-3ab(a+b)`
`=a^3 + b^3+3ab ( a + b )- 3ab ( a + b )`
`=a^3 + b^3=VT(dpcm)`
b) Ta có
`VT=a^3+b^3+c^3−3abc`
`=(a+b)^3−3ab(a+b)+c^3−3abc`
`=[(a+b)^3+c^3]−3ab(a+b+c)`
`=(a+b+c)[(a+b)^2+c^2−c(a+b)]−3ab(a+b+c)`
`=(a+b+c)(a^2+b^2+2ab+c^2−ac−bc−3ab)`
`=(a+b+c)(a^2+b^2+c^2−ab−bc−ca)=VP`
![](https://rs.olm.vn/images/avt/0.png?1311)
Biến đổi vế trái:
a + b + c 3 = a + + c 3 = a + b 3 +3 a + b 2 c+3(a+b) c 2 + c 3
= a 3 + 3 a 2 b + 3a b 2 + b 3 + 3( a 2 + 2ab + b 2 )c + 3a c 2 + 3b c 2 + c 3
= a 3 + 3 a 2 b + 3a b 2 + b 3 + 3 a 2 c + 6abc + 3 b 2 c + 3a c 2 + 3b c 2 + c3
= a 3 + b 3 + c 3 + 3 a 2 b + 3a b 2 + 3 a 2 c + 6abc + 3 b 2 c + 3a c 2 + 3b c 2
= a 3 + b 3 + c 3 + (3 a 2 b + 3a b 2 ) +( 3 a 2 c + 3abc)+ (3abc + 3 b 2 c)+(3a c 2 + 3b c 2 )
= a 3 + b 3 + c 3 + 3ab(a + b) + 3ac(a + b) + 3bc(a + b) + 3 c 2 (a + b)
= a 3 + b 3 + c 3 + 3(a + b)(ab + ac + bc + c 2 )
= a 3 + b 3 + c 3 + 3(a + b)[a(b + c) + c(b + c)]
= a 3 + b 3 + c 3 + 3(a + b)(b + c)(a + c) (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(a+b+c\right)^2+a^2+b^2+c^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2\)
\(=a^2+2ab+b^2+b^2+2bc+c^2+c^2+2ca+a^2\)
\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)
b) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left(b+c\right)\left[\left(a+b+c\right)^2+\left(a+b+c\right)a+a^2\right]-\left(b+c\right)\left(b^2+bc+c^2\right)\)
\(=\left(b+c\right)\left(3a^2+3ab+3bc+3ac\right)\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
\(a^3+b^3+c^3=3abc\\ \Leftrightarrow a^3+b^3+c^3-3abc=0\\ \Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\\ \Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\\ \Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\left(1\right)\end{matrix}\right.\\ \left(1\right)\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\\ \Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Leftrightarrow a=b=c\)
Vậy \(a^3+b^3+c^3=3abc\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
Xét vế trái:
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+3a^2bc+3abc^2+c^3-a^3-b^3-c^3\)
\(=a^3+b^3+3ab\left(a+b\right)+3\left(a+b\right)^2c+3\left(a+b\right)c^2-a^3-b^3\)
\(=3ab\left(a+b\right)+3\left(a+b\right)^2c+3\left(a+b\right)c^2\)
\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Vậy: \(\left(a+b+c\right)^3-a^3-b^3-c^3=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
(Nhớ k cho mình với nhá!)