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Ta có \(VT=\left(a+b+c\right)^3=\left[\left(a+b\right)+c\right]^3=\left(a+b\right)^3+3\left(a+b\right)^2.c+3\left(a+b\right)c^2+c^3\)
\(=a^3+3a^2b+3ab^2+b^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left[\left(a+b\right)c+c^2+ab\right]\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left[a\left(b+c\right)\right]+c\left(b+c\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=VP\)
Vậy \(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
(a+b+c)^3=((a+b)+c)^3=(a+b)^3+c^3+3(a+b)c(a+b+c)
=a^3+b^3+3ab(a+b)+c^3+3(a+b)c(a+b+c)
=a^3+b^3+c^3+3(a+b)(ab+c(a+b+c))
=a^3+b^3+c^3+3(a+b)(ab+ac+bc+c^2)
=a^3+b^3+c^3+3(a+b)(a+c)(b+c)
(a+b+c)^3
=(a+b)^3+3(a+b)^2c+3(a+b)c^2+c^3
=a^3+3a^2b+3ab^2+b^3+3(a^2+2ab+b^2)c+3(a+b)c^2+c^3
=a^3+b^3+c^3+3a^2c+6abc+3b^2c+3ac^2+3bc^2
=a^3+b^3+c^3+(3a^2c+3abc)+(3abc+3b^2c)+(3ac^2+3bc^2)
=a^3+b^3+c^3+3ac(a+b)+3bc(a+b)+3c^2(a+b)
=a^3+b^3+c^3+3(a+b)(ac+bc+c^2)
=a^3+b^3+c^3+3(a+b)[(ac+bc)+c^2]
=a^3+b^3+c^3+3(a+b)c(a+b+c)
Ta có:
\(\left(a+b+c\right)^3\)
= \(\left(a+b\right)^3+c^3+3c\left(a+b\right)\left(a+b+c\right)\)
= \(a^3+b^3+3ab\left(a+b\right)+c^3+3c\left(a+b\right)\left(a+b+c\right)\)
= \(a^3+b^3+3ab\left(a+b\right)+c^3+\left(3ac+3bc+3c^2\right)\left(a+b\right)\)
= \(a^3+b^3+c^3+\left(a+b\right)\left(3ab+3ac+3bc+3c^2\right)\)
= \(a^3+b^3+c^3+\left(a+b\right)[\left(3ab+3ac)+(3bc+3c^2\right)]\)
= \(a^3+b^3+c^3+\left(a+b\right)[3a\left(b+c)+3c(b+c\right)]\)
= \(a^3+b^3+c^3+\left(a+b\right)[\left(b+c\right)\left(3a+3c\right)]\)
= \(a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
(a+B+c)3=[(a+b)+C]3=(a+b)3+3(a+b)2c+3(a+b)c2+c3=a3+b3+3a2b+3ab2+3a2c+6abc+3b2c+3ac2+3bc2+c3
a3+b3+c3+3(a+b)(b+c)(c+a)=a3+b3+c3+6abc+3a2b+3ab2+3a2c+3b2c
+3ac2+3bc2.(nhân các đa thức 3(a+b)(a+c)(b+c) lại với nhau)
vậy (a+b+c)3=a3+b3+c3+3(a+b)(a+c)(b+c)
VT = (a+b+c)3-a3-b3-c3
= \([\left(a+b\right)+c]^3\)- a3-b3-c3
= (a+b)3+c3 +3ab(a+b)+3c(a+b)(a+b+c)-a3-b3-c3
=3(a+b) \([ab+c\left(a+b+c\right)]\)
= 3(a+b) \([ab+ac+bc+c^2]\)
= 3(a+b)(b+c)(c+a)
\(\Rightarrow\)VT=VP= 3(a+b)(b+c)(c+a)
Giải:
Ta có: \(VT=\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b+c\right)^3-a^3\right]-\left(b^3+c^3\right)\)
\(=\left(b+c\right)\left[\left(a+b+c\right)^2+\left(a+b+c\right)a+a^2\right]-\left(b+c\right)\left(b^2-bc+c^2\right)\)
\(=\left(b+c\right)\left(3a^2+3ab+3bc+3ca\right)\)
\(=3\left(b+c\right)\left[a\left(a+b\right)+c\left(a+b\right)\right]\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)=VP\) (Đpcm)
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)