a) \({\left( {a - b} \right)^3} = - {\left( {b - a} \right)^3}\)
Biến đổi vế phải thành vế trái:
\(\eqalign{
& - {\left( {b - a} \right)^3} = - ({b^3} - 3{b^2}a + 3b{a^2} - {a^3}) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\; = - {b^3} + 3{b^2}a - 3b{a^2} + {a^3} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \;= {a^3} - 3{a^2}b + 3a{b^2} - {b^3} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\; = {\left( {a - b} \right)^3} \cr} \)
Vậy \({\left( {a - b} \right)^3} = - {\left( {b - a} \right)^3}\)
Cách 2: Sử dụng quy tắc dấu ngoặc
\(\eqalign{
& {\left( {a - b} \right)^3} = {\left[ { - \left( {b{\rm{ }}-{\rm{ }}a} \right)} \right]^3} = {\left[ {\left( { - 1} \right).\left( {b{\rm{ }}-{\rm{ }}a} \right)} \right]^3} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( { - 1} \right)^3}.{\left( {b - a} \right)^3} = \left( { - 1} \right).{\left( {b - a} \right)^3} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - {\left( {b - a} \right)^3} \cr} \)
b) \({\left( { - a - b} \right)^2} = {\left( {a + b} \right)^2}\)
Biến đổi vế trái thành vế phải:
\(\eqalign{
& {\left( { - a - b} \right)^2} = {\left[ {\left( { - a} \right) + \left( { - b} \right)} \right]^2} \cr
& = {\left( { - a} \right)^2} + 2.\left( { - a} \right).\left( { - b} \right) + {\left( { - b} \right)^2} \cr
& = {a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2} \cr} \)
Vậy \({\left( { - a - b} \right)^2} = {\left( {a + b} \right)^2}\)
Cách 2: Sử dụng quy tắc dấu ngoặc
\(\eqalign{
& {\left( { - a - b} \right)^2} = {\left[ { - \left( {a + b} \right)} \right]^2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\; = {\left[ {\left( { - 1} \right).\left( {a + b} \right)} \right]^2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \;= {\left( { - 1} \right)^2}.{\left( {a + b} \right)^2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \;= 1.{\left( {a + b} \right)^2} = {\left( {a + b} \right)^2} \cr} \)
