Đặt x = a+b , y = b+c , z = c+a
Thì biểu thức trên trở thành \(x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3xy-3xyz\)
\(=\left(x+y+z\right)\left(x^2+y^2+2xy-xz-yz+z^2\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
Từ đó thay a,b,c vào rồi rút gọn :)
Rút gọn biểu thức sau: \(\left(a+b\right)^3+\left(b+c\right)^3+\left(c+a\right)^3-3\left(a+b\right)\left(b+c\right)+\left(c+a\right)\)
Rút gọn biểu thức\(\left(a+b\right)^3+\left(b+c\right)^3+\left(c+a\right)^3-3\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
Rút gọn phân thức:
\(A=\frac{\left(b-c\right)^3+\left(c-a\right)^3+\left(a-b\right)^3}{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}\)
Rút gọn
a,\(\left(a+b+c\right)^3-\left(b+c-a\right)^3-\left(b+a-c\right)^3-\left(a+c-b\right)^3\)
b,\(\left(a+b\right)^3+\left(b+c\right)^3+\left(c+a\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Rút gọn biểu thức:
\(A=\left(a+b+c\right)^3+\left(a-b-c\right)^3-6a\left(b+c\right)^2\)
Rút gọn phân thức :
\(\frac{a^3\left(b^2-c^2\right)+b^3\left(c^2-a^2\right)+c^3\left(a^2-b^2\right)}{a^2\left(b-c\right)-b^2\left(c-a\right)+c^2\left(a-b\right)}\)
Rút gọn:\(\frac{\left(b-c\right)^{^3}+\left(c-a\right)^{^3}+\left(a-b\right)^{^3}}{a^2.\left(b-c\right)+b^2.\left(a-c\right)+c^{^2}\left(a-b\right)}\)
\(\text{Rút gọn (a+b)}^3+\left(b+c\right)^3+\left(c+a\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Rút gọn
\(\frac{a^3}{\left(a-b\right)\left(a-c\right)}+\frac{b^3}{\left(b-c\right)\left(b-a\right)}+\frac{c^3}{\left(c-a\right)\left(c-b\right)}\)
