Những hằng đẳng thức đáng nhớ (tổng, hiệu 2 lập phương)

Ta có

$ \begin{array}{l} {{x}^{3}}-{{\left( 2x+1 \right)}^{3}}=x+1 \\ \Leftrightarrow \left( x-2x-1 \right)\left[ {{x}^{2}}+x\left( 2x+1 \right)+{{\left( 2x+1 \right)}^{2}}+1 \right]=0 \\ \Leftrightarrow -x-1=0 \\ \Leftrightarrow x=-1 \end{array} $

Ta có

$ \begin{array}{l} T={{\left( x-y \right)}^{3}}+{{\left( y+x \right)}^{3}}+{{\left( y-x \right)}^{3}}-3xy\left( x+y \right) \\ ={{x}^{3}}-3{{x}^{2}}y+3x{{y}^{2}}-{{y}^{3}}+{{y}^{3}}+3x{{y}^{2}}+3{{x}^{2}}y+{{x}^{3}}+{{y}^{3}}-3x{{y}^{2}}+3{{x}^{2}}y-{{x}^{3}}-3{{x}^{2}}y-3x{{y}^{2}} \\ ={{x}^{3}}+{{y}^{3}} \end{array} $

Ta có

$ \begin{array}{l} {{a}^{3}}+{{b}^{3}} \\ =\left( {{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}} \right)-\left( 3{{a}^{2}}b+3a{{b}^{2}} \right) \\ ={{\left( a+b \right)}^{3}}-3ab\left( a+b \right) \end{array} $

Vậy với $ a+b=1\Rightarrow {{a}^{3}}+{{b}^{3}}=1-3ab $

Ta có

$ \begin{array}{l} {{a}^{3}}-{{b}^{3}} \\ =\left( {{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}} \right)+\left( 3{{a}^{2}}b-3a{{b}^{2}} \right) \\ ={{\left( a-b \right)}^{3}}+3ab\left( a-b \right) \end{array} $

Vậy với $ a-b=1\Rightarrow {{a}^{3}}-{{b}^{3}}=1+3ab $

Ta có $ M=(2x+3)(4{{x}^{2}}-6x+9)-4(2{{x}^{3}}-3) $

$ =(2x+3)[{{(2x)}^{2}}-2x.3+{{3}^{2}}]-8{{x}^{3}}+12 $

$ ={{(2x)}^{3}}+{{3}^{3}}-8{{x}^{3}}+12=8{{x}^{3}}+27-8{{x}^{3}}+12=39 $ .

Vậy giá trị của $ M $ là một số lẻ.

Ta có $ {{(a+b)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}={{a}^{3}}+{{b}^{3}}+3ab(a+b)\Rightarrow {{a}^{3}}+{{b}^{3}}={{(a+b)}^{3}}-3ab(a+b) $

Từ đó $ B={{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc={{(a+b)}^{3}}-3ab(a+b)+{{c}^{3}}-3abc $

$ =\left[ {{(a+b)}^{3}}+{{c}^{3}} \right]-3ab(a+b+c) $

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

Mà $ a+b+c=0 $ nên $ B=0.\left[ {{(a+b)}^{2}}-(a+b)c+{{c}^{2}} \right]-3ab.0=0 $

Vậy $ B=0 $ .

Ta có $ A={{x}^{3}}-3{{x}^{2}}+3x={{x}^{3}}-3{{x}^{2}}+3x-1+1={{(x-1)}^{3}}+1 $

Thay $ x=1001 $ vào $ A={{(x-1)}^{3}}+1 $ ta được $ A={{(1001-1)}^{3}}+1 $ suy ra $ A={{1000}^{3}}+1 $ .

Ta có $ (x+2)({{x}^{2}}-2x+4)-x({{x}^{2}}-2)=14 $

$ \Leftrightarrow {{x}^{3}}+{{2}^{3}}-({{x}^{3}}-2x)=14\Leftrightarrow {{x}^{3}}+8-{{x}^{3}}+2x=14 $

$ \Leftrightarrow 2x=6\Leftrightarrow x=3 $ .

Vậy $ x=3 $ .

Ta có

$ \begin{array}{l} \left( {{x}^{2}}+3 \right)\left( {{x}^{4}}+9-3{{x}^{2}} \right)-4{{x}^{3}}-23 \\ ={{\left( {{x}^{2}} \right)}^{3}}+{{3}^{3}}-4{{x}^{3}}-23 \\ ={{x}^{6}}-4{{x}^{3}}+4 \\ ={{\left( {{x}^{3}} \right)}^{2}}-2.{{x}^{3}}.2+{{2}^{2}} \\ ={{\left( {{x}^{3}}-2 \right)}^{2}} \end{array} $

Ta có

$ \begin{array}{l} \left( x+3 \right)\left( {{x}^{2}}-3x+9 \right)-x\left( x-2 \right)\left( x+2 \right)=15 \\ \Leftrightarrow {{x}^{3}}+27-x\left( {{x}^{2}}-4 \right)=15 \\ \Leftrightarrow {{x}^{3}}+27-{{x}^{3}}+4x=15 \\ \Leftrightarrow 4x=-12 \\ \Leftrightarrow x=-3 \end{array} $

Ta có $ {{x}^{3}}+3{{x}^{2}}+3x+1=0\Leftrightarrow {{(x+1)}^{3}}=0 $

$ \Leftrightarrow x+1=0\Leftrightarrow x=-1 $ .

Vậy $ x=-1 $ .

$ \left( 3x+2y \right)\left( 9{{x}^{2}}-6xy+4{{y}^{2}} \right)-\left( 3x-2y \right)\left( 9{{x}^{2}}+6xy+4{{y}^{2}} \right) $

$ =\left( 3x+2y \right)\left[ {{\left( 3x \right)}^{2}}-3x.2y+{{\left( 2y \right)}^{2}} \right]-\left( 3x-2y \right)\left[ {{\left( 3x \right)}^{2}}+3x.2y+{{\left( 2y \right)}^{2}} \right] $

$ =\left[ {{\left( 3x \right)}^{3}}+{{\left( 2y \right)}^{3}} \right]-\left[ {{\left( 3x \right)}^{3}}-{{\left( 2y \right)}^{3}} \right] $

$ ={{\left( 3x \right)}^{3}}+{{\left( 2y \right)}^{3}}-{{\left( 3x \right)}^{3}}+{{\left( 2y \right)}^{3}}=16{{y}^{3}} $

Ta có

$ \begin{array}{l} 27{{x}^{3}}-{{a}^{3}}{{b}^{3}} \\ ={{\left( 3x \right)}^{3}}-{{\left( ab \right)}^{3}} \\ =\left( 3x-ab \right)\left( 9{{x}^{2}}+3xab+{{a}^{2}}{{b}^{2}} \right) \end{array} $

Ta có $ E=(x+1)({{x}^{2}}-x+1)-(x-1)({{x}^{2}}+x+1)={{x}^{3}}+1-({{x}^{3}}-1)={{x}^{3}}+1-{{x}^{3}}+1=2 $

Vậy $ E=2 $ .

Ta có

$ \begin{array}{l} \left( x+3 \right)\left( {{x}^{2}}-3x+9 \right)-\left( 30+2x+{{x}^{3}} \right) \\ =\left( x+3 \right)\left( {{x}^{2}}-3x+{{3}^{2}} \right)-\left( 30+2x+{{x}^{3}} \right) \end{array} $

$ ={{x}^{3}}+{{3}^{3}}-\left( 30+2x+{{x}^{3}} \right) $

$ =-3-2x $