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:
\(\left(a+b+c\right)^3-\left(a+b-c\right)^3-\left(b+c-a\right)^3+\left(c+a-b\right)^3\)
\(=3\left(a+b-c+b+c-a\right)\left(b+c-a+c+a-b\right)\left(c+a-b+a+b-c\right)\)
\(=3.2b.2c.2a=24abc\)
