Bài 7. Lập phương của một tổng. Lập phương của một hiệu

\(\left(a+b\right)^3=\left(a+b\right)^2\cdot\left(a+b\right)\)

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

\(=a^3+a^2b+2a^2b+2ab^2+ab^2+b^3\)

\(=a^3+3a^2b+3ab^2+b^3\)

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

1. 

a)      \({\left( {x + 3} \right)^3} = {x^3} + 3.{x^2}.3 + 3.x{.3^2} + {3^3} = {x^3} + 9{x^2} + 27x + 27\)

b)      \({\left( {x + 2y} \right)^3} = {x^3} + 3.{x^2}.2y + 3.x.{\left( {2y} \right)^2} + {\left( {3y} \right)^3} = {x^3} + 6{x^2}y + 12x{y^2} + 27{y^3}\)

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

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

Từ đó ta có \({\left( {a - b} \right)^3} = {a^3} - 3{a^2}b + 3a{b^2} - {b^3}\)

\(\left(2x-y\right)^3=8x^3-12x^2y+6xy^2-y^3.\)

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

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

a)

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

b)

\({\left( {\dfrac{1}{2}x - 1} \right)^3} = {\left( {\dfrac{1}{2}x} \right)^3} - 3.{\left( {\dfrac{1}{2}x} \right)^2}.1 + 3.\dfrac{1}{2}x{.1^2} - {1^3} = \dfrac{1}{8}{x^3} - \dfrac{3}{4}{x^2} + \dfrac{3}{2}x - 1\)

a)      \(27 + 54x + 36{x^2} + 8{x^3} = {3^3} + {3.3^2}.2x + 3.3.{\left( {2x} \right)^2} + {\left( {2x} \right)^3} = {\left( {3 + 2x} \right)^3}\)

b)      \(64{x^3} - 144{x^2}y + 108x{y^2} - 27{y^3} = {\left( {4x} \right)^3} - 3.{\left( {4x} \right)^2}.3y + 3.4x.{\left( {3y} \right)^2} - {\left( {3y} \right)^3} = {\left( {4x - 3y} \right)^3}\)