Giải bài 2 trang 39 vở thực hành Toán 8

Đề bài

Phân tích các đa thức sau thành nhân tử:

a) \({x^3}\; + {y^3}\; + x + y\);

b) \({x^3}\;-{y^3}\; + x-y\);

c) \({\left( {x-y} \right)^3}\; + {\left( {x + y} \right)^3}\);

d) \({x^3}\;-3{x^2}y + 3x{y^2}\;-{y^3}\; + {y^2}\;-{x^2}\).

Lời giải chi tiết

a) Ta có \({x^3}\; + {y^3}\; + x + y = \left( {{x^3}\; + {y^3}} \right) + \left( {x + y} \right)\)

\(\begin{array}{*{20}{l}}{ = \left( {x + y} \right)\left( {{x^2}\;-xy + {y^2}} \right) + \left( {x + y} \right)}\\{ = \left( {x + y} \right)\left( {{x^2}\;-xy + {y^2}\; + 1} \right).}\end{array}\)

b) Ta có \({x^3}\;-{y^3}\; + x-y = \left( {{x^3}\;-{y^3}} \right) + \left( {x-y} \right)\)

\(\begin{array}{*{20}{l}}{ = \left( {x-y} \right)\left( {{x^2}\; + xy + {y^2}} \right) + \left( {x-y} \right)}\\{ = \left( {x-y} \right)\left( {{x^2}\; + xy + {y^2}\; + 1} \right).}\end{array}\)

c) Ta có \({\left( {x-y} \right)^3}\; + {\left( {x + y} \right)^3}\; = \left( {x-y + x + y} \right).\left[ {{{\left( {x + y} \right)}^2}\;-\left( {x + y} \right)\left( {x - y} \right) + {{\left( {x - y} \right)}^2}} \right]\)

\(\begin{array}{*{20}{l}}{ = \;2x.\left[ {{x^2}\; + 2xy + {y^2}\;-\left( {{x^2}\;-{y^2}} \right) + {x^2}\; - 2xy + {y^2}} \right]}\\{ = \;2x.\left[ {\left( {{x^2}\;-{x^2}\; + {x^2}} \right)\; + \;\left( {2xy - 2xy} \right)\; + \;\left( {{y^2}\; + {y^2}\; + {y^2}} \right)} \right]}\\{ = 2x\left( {{x^2}\; + 3{y^2}} \right).}\end{array}\)

d) Ta có \({x^3}\;-3{x^2}y + 3x{y^2}\;-{y^3}\; + {y^2}\;-{x^2}\; = \left( {{x^3}\;-3{x^2}y + 3x{y^2}\;-{y^3}} \right)-\left( {{x^{2\;}}-{y^2}} \right)\)

\(\begin{array}{*{20}{l}}{ = {{\left( {x-y} \right)}^3}\;-\left( {x-y} \right)\left( {x + y} \right)}\\{ = \left( {x-y} \right).\left[ {{{\left( {x-y} \right)}^{2\;}}-\left( {x + y} \right)} \right]}\\{ = \left( {x-y} \right)\left( {{x^2}\;-2xy + {y^{2\;}}-x-y} \right).}\end{array}\)