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

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

\(=\left(b+c\right)\left(3a^3+3ab+3bc+3ca\right)=3\left(b+c\right)\left[a\left(a+b\right)+c\left(a+b\right)\right]\)

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

Bai lam 

\(\frac{x^3-8}{x^2+2x+4}=\frac{\left(x-2\right)\left(x^2+2x+4\right)}{x^2+2x+4}=x-2\)

Hoc tot 

\(x^3-8\div x^2+2x+4\)

\(=\frac{x^3-8}{x^2+2x+4}\)

\(=\frac{\left(x-2\right)\left(x^2+2x+4\right)}{x^2+2x+4}\)

\(=x-2\)

Bai 1 

\(x^2+x-30=x^2+6x-5x-30=\left(x-5\right)\left(x+6\right)\)

Bai 2 

a, \(\left(x-2\right)^2-x\left(x-5\right)=13\)

\(\Leftrightarrow x^2-4x+4-x^2+5x=13\)

\(\Leftrightarrow x+4=13\Leftrightarrow x=9\)

b, \(4x^3-100x=0\Leftrightarrow x\left(4x^2-100\right)=0\)

\(\Leftrightarrow x\left(2x-10\right)\left(2x+10\right)=0\Leftrightarrow x=0;\pm5\)

\(\frac{4}{x-5}\)- \(\frac{1}{x+5}\)+\(\frac{13x-x^2}{\left(5-x\right)\left(5+x\right)}\)= \(\frac{4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}\)- \(\frac{x-5}{\left(x-5\right)\left(x+5\right)}\)- \(\frac{x\left(x-13\right)}{\left(x-5\right)\left(5+x\right)}\)=\(\frac{4\left(x+5\right)-x+5+x\left(x-13\right)}{\left(x-5\right)\left(x+5\right)}\)= \(\frac{x^2-10x+12}{\left(x-5\right)\left(x+5\right)}\)

\(\frac{4}{x-5}-\frac{1}{x+5}+\frac{13x-x^2}{\left(5-x\right)\left(x+5\right)}\)ĐKXĐ : \(x\ne\pm5\)

\(=\frac{4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{\left(x+5\right)\left(x-5\right)}-\frac{13x-x^2}{\left(x-5\right)\left(x+5\right)}\)

\(=\frac{4x+20-x+5-13x+x^2}{\left(x-5\right)\left(x+5\right)}=\frac{-10x+25+x^2}{\left(x-5\right)\left(x+5\right)}\)

\(=\frac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}=\frac{x-5}{x+5}\)