\(a^3+b^3+c^3=3abc\)

\(\Rightarrow a^3+b^3+c^3-3abc=0\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{matrix}\right.\)

+) \(a^2+b^2+c^2-ab-bc-ca=0\)

\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Mà \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{matrix}\right.\Rightarrow a=b=c\)

\(\Rightarrowđpcm\)

\(VT=x\left(x+1\right)\left(x+2\right)=\left(x^2+x\right)\left(x+2\right)\)

\(=x^3+3x^2+2x=VP\)

\(\Rightarrowđpcm\)

\(VT=x\left(x+1\right)\left(x+2\right)\)

\(=\left(x^2+1\right)\left(x+2\right)\)

\(=x^3+3x^2+2x=VP\)

\(\Rightarrowđpcm\)

Ta có: \(3^3\equiv1\left(mod13\right)\)

\(\Rightarrow3^{15}\equiv1\left(mod13\right)\)

\(9^3\equiv1\left(mod13\right)\)

\(\Rightarrow9^6\equiv1\left(mod13\right)\)

\(\Rightarrow3^{15}-9^6\equiv0\left(mod13\right)\)

\(\Rightarrow3^{15}-9^6⋮13\)

\(\Rightarrowđpcm\)

a chia hết cho m => a =mq

b không chia hết cho m => b =m.p+r  với   r < m

=>a+b =mq+mp+r =m(q+p) +r  => a+b khoog chia hết cho m

Bài 1:

\(a,\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\)

\(=2x\left(4x+2\right)=4x\left(2x+1\right)\)

b, \(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3+z^3-3xyz-3xy\left(x+y\right)\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-yz-xz+z^2\right)-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)

\(2^{x+2}-2^x=96\)

\(\Rightarrow2^x.4-2^x=96\)

\(\Rightarrow2^x\left(4-1\right)=96\)

\(\Rightarrow2^x=32\)

\(\Rightarrow x=5\)

Vậy x = 5

\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)

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

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

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

\(=ab\left(a-b\right)-\left(a-b\right)\left(ac+bc\right)+c^2\left(a-b\right)\)

\(=\left(a-b\right)\left(ab-ac-bc+c^2\right)\)

\(=\left(a-b\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]\)

\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)

\(ab\left(x^2+1\right)+x\left(a^2+b^2\right)\)

\(=abx^2+ab+a^2x+b^2x\)

\(=ax\left(bx+a\right)+b\left(a+bc\right)\)

\(=\left(ax+b\right)\left(bx+a\right)\)