Bài 32:
2: Thay x=1 vào nhị thức, ta được:
\(\left(5\cdot1-6\right)^{2021}=\left(-1\right)^{2021}=-1\)
=>Tổng các hệ số của tất cả các hạng tử sau khi khai triển là -1
Bài 33:
1:
Đặt m=a+b-2c; n=b+c-2a; i=c+a-2b
=>m+n+i=a+b-2c+b+c-2a+c+a-2b=0
=>m+n=-i
Đặt X=\(\left(a+b-2c\right)^3+\left(b+c-2a\right)^3+\left(c+a-2b\right)^3\)
\(=m^3+n^3+i^3\)
\(=\left(m+n\right)^3-3mn\left(m+n\right)+i^3\)
\(=\left(-i\right)^3-3mn\cdot\left(-i\right)-i^3=3mni\)
=3(a+b-2c)(b+c-2a)(c+a-2b)
2: \(\left(a+b+c\right)^3+\left(a-b-c\right)^3\)
\(=a_{}^3+3a^2\left(b+c\right)+3a\left(b+c\right)^2+\left(b+c\right)^3+a^3-3a^2\left(b+c\right)+3a\left(b+c\right)^2-\left(b+c\right)^3\)
\(=2a^3+6a\left(b+c\right)^2=2a\left\lbrack a^2+3\left(b+c\right)^2\right\rbrack\)
\(\left(b-c-a\right)^3+\left(c-a-b\right)^3\)
\(=\left(-a+b-c\right)^3+\left(-a-b+c\right)^3\)
\(=-\left(a-b+c\right)^3-\left(a+b-c\right)^3\)
\(=-\left\lbrack a^3-3a^2\left(b-c\right)+3a\left(b-c\right)^2-\left(b-c\right)^3+a^3+3a^2\left(b-c\right)+3a\left(b-c\right)^2+\left(b-c\right)^3\right\rbrack\)
\(=-\left\lbrack2a^3+6a\left(b-c\right)^2\right\rbrack=-2a^3-6a\left(b-c\right)^2\)
Ta có: \(\left(a+b+c\right)^3+\left(a-b-c\right)^3+\left(b-c-a\right)^3+\left(c-a-b\right)^3\)
\(=2a^3+6a\left(b+c\right)^2-2a^3-6a\left(b-c\right)^2\)
\(=6a\left(b^2+2bc+c^2-b^2+2bc-c^2\right)=6a\cdot4bc=24abc\)
