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\(\left(a+b+c\right)^3=\left(a+b\right)^3+3\left(a+b\right)c\left(a+b+c\right)+c^3\)
\(=a^3+3ab\left(a+b\right)+b^3+3c\left(a+b\right)\left(a+b+c\right)+c^3\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]=a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)\left(\text{đ}pcm\right)\)
a) \(\left(a+b+c\right)^3-\left(b+c-a\right)^3-\left(a+c-b\right)^3-\left(a+b-c\right)^3\)
\(=\left[\left(a+b\right)+c\right]^3-\left[\left(b+c\right)-a\right]^3-\left[\left(a+c\right)-b\right]^3-\left[\left(a+b\right)-c\right]^3\)
\(=\left[\left(a+b\right)^3+3.\left(a+b\right)^2.c+3.\left(a+b\right).c^2+c^3\right]-\left[\left(b+c\right)^3-3.\left(b+c\right)^2.a+3.\left(b+c\right).a^2-a^3\right]-\left[\left(a+c\right)^3-3.\left(a+c\right)^2.b+3.\left(a+c\right).b^2-b^3\right]-\left[\left(a+b\right)^3-3.\left(a+b\right)^2.c+3.\left(a+b\right).c^2-c^3\right]\)\(=\left[\left(a^3+3a^2b+3ab^2+b^3\right)+3\left(a^2+2ab+b^2\right).c+3c^2a+3c^2b+c^3\right]-\left[\left(b^3+3b^2c+3bc^2+c^3\right)-3.\left(b^2+2bc+c^2\right).a+3a^2b+3a^2c-a^3\right]-\left[\left(a^3+3a^2c+3ac^2+c^3\right)-3\left(a^2+2ab+b^2\right).c+3c^2a+3c^2b-c^3\right]\)\(=\left(a^3+3a^2b+3ab^2+b^3+3ca^2+6abc+3b^2c+3c^2a+3c^2b+c^3\right)-\left(b^3+3b^2c+3bc^2+c^3-3ab^2-6abc-3ac^2+3a^2b+3a^2c-a^3\right)-\left(a^3+3a^2c+3ac^2+c^3-3a^2c-6abc-3b^2c+3c^2a+3c^2b-c^3\right)\)\(=a^3+3a^2b+3ab^2+b^3+3ca^2+6abc+3b^2c+3c^2a+3c^2b+c^3-b^3-3b^2c-3bc^2-c^3+3ab^2+6abc+3ac^2-3a^2b-3a^2c+a^3-a^3-3a^2c-3ac^2-c^3+3a^2c+6abc+3b^2c-3c^2a-3c^2b+c^3\)\(=3ab^2+6abc+3ab^2+6abc+a^3+6abc+3b^2c-3c^2b\)
\(=6ab^2+18abc+a^3+3b^2c-3bc^2\)
P/s: Ko chắc! Đây là kết quả của hơn 50 phút !
a. Câu hỏi của Nhàn Nguyễn - Toán lớp 8 - Học toán với OnlineMath
a) Đặt a+b-c=x , b+c-a=y, c+a-b=z
⇒(a+b+c)3−x3−y3−z3
Có x + y +z = a+b-c + b+c-a+c+a-b = a+b+c
⇒(x+y+z)3−x3−y3−z3
=[(x+y)+z3]−x3−y3−z3
=(x+y)3+z3+3z(x+y)(x+y+z)−x3−y3−z3
=x3+y3+3xy(x+y)+z3+3z(x+y)(x+y+z)−x3−y3−z3
=3(x+y)(xy+xz+yz+z2)
=3(x+y)[x(y+z)+z(y+z)]
=3(x+y)(y+z)(x+z)
Áp dụng hằng đẳng thức trên ta có
3(a+b-c+b+c-a)(b+c-a+c+a-b)(a+b-c+c+a-b)
= 3.2b.2c.2a
= 24abc
a) Ta có:
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)
\(=a^3+b^3+3ab\left(a+b\right)+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
b) Đặt a + b - c = x
b + c - a = y
c + a - b = z
=> x + y + z = a + b - c + b + c - a + c + a - b = a + b + c
Áp dụng hằng đẳng thức \(\left(x+y+z\right)^3-x^3-y^3-z^3=3\left(x+y\right)\left(y+z\right)\left(x+z\right)\) ( Câu a )
Ta có:
\(\left(a+b+c\right)^3-\left(a+b-c\right)^3-\left(b+c-a\right)^3+\left(c+a-b\right)^3\)
\(=3\left(a+b-c+b+c-a\right)\left(b+c-a+c+a-b\right)\left(c+a-b+a+b-c\right)\)
\(=3.2b.2c.2a=24abc\)
Đặt \(a+b-c=x;b+c-a=y;a+c-b=z\)
Lúc đó \(x+y+z=b+c-a+a+b-c+a+c-b=a+b+c\)
\(\Rightarrow bt=\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+3\left(x+y\right)^2z+3z^2\left(x+y\right)+z^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+3z\left(x+y\right)\left(x+y+z\right)+z^3-x^3-y^3-z^3\)
\(=x^3+3x^2y+3xy^2+y^2+3z\left(x+y\right)\left(x+y+z\right)\)
\(+z^3-x^3-y^3-z^3\)
\(=x^3+3xy\left(x+y\right)+y^2+3z\left(x+y\right)\left(x+y+z\right)\)
\(+z^3-x^3-y^3-z^3\)
\(=3xy\left(x+y\right)+3z\left(x+y\right)\left(x+y+z\right)\)
\(=3\left(x+y\right)\left(xy+xz+zy+z^2\right)\)
\(=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
\(=3\left(x+y\right)\left(x+z\right)\left(y+z\right)\)
\(1.a\left(a+2b\right)^3-b\left(2a+b\right)^3\)
=\(a\left(a^3+6a^2b+12ab^2+8b^3\right)-b\left(8a^3+12a^2b+6ab^2+b^3\right)\)
=\(a^4+6a^3b+12a^2b^2+8ab^3-8a^3b-12a^2b^2-6ab^3-b^4\)
=\(a^4-b^4\)=\(\left(a^2-b^2\right)\left(a^2+b^2\right)\)
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(=\left[\left(a+b\right)^3+c^3+3c.\left(a+b\right).\left(a+b+c\right)\right]-a^3-b^3-c^3\)
\(=\left[a^3+b^3+3ab.\left(a+b\right)+c^3+3c.\left(a+b\right)\right]-a^3-b^3-c^3\)
\(=3ab.\left(a+b\right)+3c.\left(a+b\right)\left(a+b+c\right)=3.\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Áp dụng :
Đặt \(\left\{{}\begin{matrix}a+b-c=x\\a-b+c=y\\-a+b+c=z\end{matrix}\right.\) \(\Rightarrow x+y=z=a+b+c\)
Khi đó biểu thức trở thành :
\(\left(x+y+z\right)^3-x^3-y^3-z^3=3.\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
\(=3.2a.2b.2c=24abc\)