\(\left(x+a\right)\left(x+b\right)\left(x+c\right)\)

\(=x^3+ax^2+bx^2+cx^2+abx+acx+bcx+abc\)

\(=x^3+x^2\left(a+b+c\right)+x\left(ab+ac+bc\right)+abc\)

\(=x^3+6x^2-7x-60\)

\(\left(x+a\right)\left(x+b\right)\left(x+c\right)\)

= \(\left(x^2+xb+ax+ab\right).\left(x+c\right)\)

= \(x^3+x^2c+x^2b+xbc+ax^2+axc+abx+abc\)

= \(x^3+x^2\left(a+b+c\right)+x\left(ab+ac+bc\right)+abc\)

= \(x^3+6x^2-7x-60\)

không biết làm hâhha

\(A=\left(x^2+\left(a+b\right)x+ab\right)\left(x+c\right)=x^3+\left(a+b+c\right)x^2+\left(ab+bc+ac\right)x+abc\)

\(A=x^3+6x^2-7x-60\)

Nếu rút gọn thành nhân tử thì:

\(A=x^3-3x^2+9x^2-27x+20x-60=x^2\left(x-3\right)+9x\left(x-3\right)+20\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2+9x+20\right)=\left(x-3\right)\left(x^2+4x+5x+20\right)=\left(x-3\right)\left[x\left(x+4\right)+5\left(x+4\right)\right]\)

\(A=\left(x-3\right)\left(x+4\right)\left(x+5\right)\).

\(A=\left(x+a\right)\left(x+b\right)\left(x+c\right)\)

    \(=\left(x^2+ax+bx+ab\right)\left(x+c\right)\)

    \(=x^3+ax^2+bx^2+abx+cx^2+acx+bcx+abc\)

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

Theo bài ra ta có:

\(a+b+c=6\)

\(ab+bc+ca=-7\)

\(abc=-60\)

\(\Rightarrow A=x^3+6x^2-7x-60\)

\(A=\left(x+a\right)\left(x+b\right)\left(x+c\right)\\ =\left(x^2+ax+bx+ab\right)\left(x+c\right)\\ =x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc\\ =x^3+6x^2-7x-60\)

\(B=\left(x+y+z\right)^2=\left[\left(x+y\right)+z\right]^2\\ =\left(x+y\right)^2+2\left(x+y\right)z+z^2\\ =x^2+2xy+y^2+2xz+2yz+z^2\\ =x^2+y^2+z^2+2xy+2yz+2zx\)

(x+a)(x+b)(x+c)=x^3+Bx^2+Cx+D
a,b,c là nghiệm PT nên
[a+b+c=-B=6=>B=-6
[ab+bc+ca=C=-7
[abc = -D=-60=>D=60
tổng hệ số của đa thức:
1+B+C+D=1-6-7+60=48