Question

Cho (a+b+c)^3=a^3+b^3+c^3      tính giá trị T =(a+b)(b+c)^2(c+a)^2

Answer 1

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

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

=>\(3\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)

=>\(\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)

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

\(=\left(a+b\right)\left(a+c\right)\left(b+c\right)\cdot\left(b+c\right)\left(a+c\right)\)

\(=0\cdot\left(b+c\right)\left(a+c\right)\)

=0