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

Ta có

$ \begin{array}{l} -27{{x}^{3}}~+27{{x}^{2}}-9x+1 \\ ={{\left( -3x \right)}^{3}}+3.{{\left( -3x \right)}^{2}}.1+3\left( -3x \right).1+{{1}^{3}} \\ ={{\left( -3x+1 \right)}^{3}} \end{array} $

Ta có

$ \begin{array}{l} {{\left( 2{{x}^{2}}+y \right)}^{3}} \\ ={{\left( 2{{x}^{2}} \right)}^{3}}+3.{{\left( 2{{x}^{2}} \right)}^{2}}.y+3.2{{x}^{2}}.{{y}^{2}}+{{y}^{3}} \\ =8{{x}^{6}}+12{{x}^{4}}y+6{{x}^{2}}{{y}^{2}}+{{y}^{3}} \end{array} $

Ta có

$ \begin{array}{l} 8{{x}^{3}}~-36{{x}^{2}}y+54x{{y}^{2}}-27{{y}^{3}} \\ ={{\left( 2x \right)}^{3}}+3.{{\left( 2x \right)}^{2}}.\left( -3y \right)+3\left( x \right){{\left( -3y \right)}^{2}}-27{{y}^{3}} \\ ={{\left( 2x-3y \right)}^{3}} \end{array} $

Ta có $ {{(-b-a)}^{3}}={{\text{ }\!\![\!\!\text{ }-(a+b)\text{ }\!\!]\!\!\text{ }}^{3}}=-{{(a+b)}^{3}} $

$ =-({{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}})=-{{a}^{3}}-3{{a}^{2}}b-3a{{b}^{2}}-{{b}^{3}} $

$ =-{{a}^{3}}-3ab(a+b)-{{b}^{3}} $ .

$ {{(c-d)}^{3}}={{c}^{3}}-3{{c}^{2}}d+3c{{d}^{2}}+{{d}^{3}}={{c}^{3}}-{{d}^{3}}+3cd(d-c) $

$ {{(y-1)}^{3}}={{y}^{3}}-3{{y}^{2}}.1+3y{{.1}^{2}}-{{1}^{3}}={{y}^{3}}-1-3y(y-1) $

$ {{(y-2)}^{3}}={{y}^{3}}-3{{y}^{2}}.2+3y{{.2}^{2}}-{{2}^{3}}={{y}^{3}}-6{{y}^{2}}+12y-8={{y}^{3}}-8-6y(y-2)\ne {{y}^{3}}-8-6y(y+2) $ .

Ta có

$ \begin{array}{l} T={{\left( 2x \right)}^{3}}-3.{{\left( 2x \right)}^{2}}.{{y}^{2}}+3.2x.{{\left( {{y}^{2}} \right)}^{2}}-{{\left( {{y}^{2}} \right)}^{3}} \\ ={{\left( 2x-{{y}^{2}} \right)}^{3}} \end{array} $

Khi đó với $ x=3;y=2 $ ta được $ T={{\left( 2.3-4 \right)}^{3}}={{2}^{3}}=8 $

Ta có

$ \begin{array}{l} {{\left( x-y+1 \right)}^{3}} \\ ={{\left( x-y \right)}^{3}}+3{{\left( x-y \right)}^{2}}+3\left( x-y \right)+1 \\ ={{x}^{3}}-3{{x}^{2}}y+3x{{y}^{2}}-{{y}^{3}}+3{{x}^{2}}-6xy+3{{y}^{2}}+3x-3y+1 \end{array} $

Ta có

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

Khi đó với $ x=10 $ ta được $ T={{\left( 10-4 \right)}^{3}}={{6}^{3}}=216 $

Ta có

$ \begin{array}{l} {{x}^{3}}-15{{x}^{2}}+75x-133=0 \\ \Leftrightarrow {{x}^{3}}-3.{{x}^{2}}.5+3.x{{.5}^{2}}-{{5}^{3}}-8=0 \\ \Leftrightarrow {{\left( x-5 \right)}^{3}}=8 \\ \Leftrightarrow x-5=2 \\ \Leftrightarrow x=7 \end{array} $

$\begin{array}{*{20}{l}} {{{\left( {\dfrac{1}{2}x - 2} \right)}^3}}\\ { = {{\left( {\dfrac{x}{2}} \right)}^3} - 3.{{\left( {\dfrac{x}{2}} \right)}^2}.2 + 3.\dfrac{x}{2}{{.2}^2} - {2^3}}\\ { = \dfrac{{{x^3}}}{8} - \dfrac{{3{x^2}}}{2} + 6x - 8} \end{array}$

Ta có

$ \begin{array}{l} {{\left( x-3y \right)}^{3}}={{x}^{3}}-9{{x}^{2}}y+27x{{y}^{2}}-27{{y}^{3}} \\ {{\left( 2x+y \right)}^{3}}=8{{x}^{3}}+12{{x}^{2}}y+6x{{y}^{2}}+{{y}^{3}} \\ \Rightarrow {{\left( x-3y \right)}^{3}}-{{\left( 2x+y \right)}^{3}}=-7{{x}^{3}}-21{{x}^{2}}y+21x{{y}^{2}}-28{{y}^{3}} \end{array} $

$ \begin{array}{l} {{\left( 4x-1 \right)}^{3}}-\left( 4x-3 \right)\left( 16{{x}^{2}}+3 \right)={{x}^{3}} \\ \Leftrightarrow 64{{x}^{3}}-48{{x}^{2}}+12x-1-64{{x}^{3}}-12x+48{{x}^{2}}+9={{x}^{3}} \\ \Leftrightarrow 8={{x}^{3}} \\ \Leftrightarrow x=2 \end{array} $

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