Question

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,\(\left(a+b\right)^3+\left(b+c\right)^3+\left(c+a\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

Answer 1

a,\(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a^2b+b^2a+c^2a+ca^2+b^2c+c^2b\right)\)

Tương tự :

\(\left(b+c-a\right)^3=b^3+c^3-a^3+3\left(a^2b-b^2a+ca^2-ac^2+b^2c+c^2b\right)\)

\(\left(b+a-c\right)^3=b^3-c^3+a^3+3\left(a^2b+b^2a-ca^2+ac^2-b^2c+c^2b\right)\)

\(\left(a+c-b\right)^3=c^3+a^3-b^3+3\left(-a^2b+b^2a+ca^2+ac^2+b^2c-c^2b\right)\)

Biểu thức sau khi rút gọn ta được 

24abc

Answer 2

b,\(\left(a+b\right)^3=a^3+b^3+3\left(a^2b+b^2a\right)\)

\(\left(c+b\right)^3=c^3+b^3+3\left(c^2b+b^2c\right)\)

\(\left(a+c\right)^3=a^3+c^3+3\left(a^2c+b^2c\right)\)

=>\(\left(a+b\right)^3+\left(b+c\right)^3+\left(c+a\right)^3=\)\(2\left(a^2+b^2+c^2\right)+3\left(a^2b+b^2a+c^2a+ca^2+b^2c+c^2b\right)\)

Lại có 

\(3\left(a+b\right)\left(b+c\right)\left(c+a\right)=\left(3\left(a^2b+b^2a+c^2a+ca^2+b^2c+c^2b+2abc\right)\right)\)

Biểu thức khi đó trở thành 

\(2\left(a^2+b^2+c^2\right)-6abc=2\left(a^2+b^2+c^2-3abc\right)\)

Tặng vk iu