2. Bài 7: Lập phương của một tổng hay một hiệu

\(\begin{array}{l}P = {\left( {4x + 1} \right)^3}\;-\left( {4x + 3} \right)\left( {16{x^2}\; + 3} \right)\\\,\,\,\,\,\,\,\begin{array}{*{20}{l}}{ = {{\left( {4x} \right)}^3}\; + 3.{{\left( {4x} \right)}^2}.1 + 3.4x{{.1}^2}\; + {1^3}\;-(64{x^3}\; + 12x + 48{x^2}\; + 9)}\\\begin{array}{l} = 64{x^3}\; + 48{x^2}\; + 12x + 1-64{x^3}\;-12x-48{x^2}\;-9\\ =  – 8\end{array}\end{array}\end{array}\)

\(\begin{array}{l}Q = {\left( {x-2} \right)^3}\;-x\left( {x + 1} \right)\left( {x-1} \right) + 6x\left( {x-3} \right) + 5x\\\,\,\,\,\,\,\,\,\begin{array}{*{20}{l}}{ = {x^3}\;-3.{x^2}.2 + 3x{{.2}^2}\;-{2^3}\;-x\left( {{x^2}\;-1} \right) + 6{x^2}\;-18x + 5x}\\\begin{array}{l} = {x^3}\;-6{x^2}\; + 12x-8-{x^3}\; + x + 6{x^2}\;-18x + 5x\\ =  – 8\end{array}\end{array}\end{array}\)

\( \Rightarrow P = Q\)

\(\begin{array}{l}P = 8{x^3}\;-12{x^2}y + 6x{y^2}\;-{y^3}\; + 12{x^2}\;-12xy + 3{y^2}\; + 6x-3y + 11\\\,\,\,\,\,\begin{array}{*{20}{l}}{ = {{\left( {2x-y} \right)}^3}\; + 3{{\left( {2x-y} \right)}^2}\; + 3\left( {2x-y} \right) + 1 + 10}\\{\; = {{\left( {2x-y + 1} \right)}^3}\; + 10}\end{array}\end{array}\)

\(\begin{array}{l}\;\,\,\,\,\,\,{(x + 1)^3} – {(x – 1)^3} – 6{(x – 1)^2} = 20\\ \Leftrightarrow {x^3} + 3{x^2} + 3x + 1 – \left( {{x^3} – 3{x^2} + 3x – 1} \right) – 6\left( {{x^2} – 2x + 1} \right) =  – 10\\ \Leftrightarrow {x^3} + 3{x^2} + 3x + 1 – {x^3} + 3{x^2} – 3x + 1 – 6{x^2} + 12x – 6 =  – 10\\ \Leftrightarrow 12x – 4 = 20\\ \Leftrightarrow 12x = 20 + 4\\ \Leftrightarrow 12x = 24\\ \Leftrightarrow x = 2\end{array}\)

\( \Rightarrow a = 2\). Vậy ước của \(2\) là \(2\).

Ta có

\(\begin{array}{l}\;P = {\left( {4x + 1} \right)^3}\;-\left( {4x + 3} \right)(16{x^2}\; + 3)\\\,\,\,\,\,\,\,\,\begin{array}{*{20}{l}}{ = {{\left( {4x} \right)}^3}\; + 3.{{\left( {4x} \right)}^2}.1 + 3.4x{{.1}^2}\; + {1^3}\;-\left( {64{x^3}\; + 12x + 48{x^2}\; + 9} \right)}\\\begin{array}{l} = 64{x^3}\; + 48{x^2}\; + 12x + 1-64{x^3}\;-12x-48{x^2}\;-9\\ =  – 8\end{array}\end{array}\\Q = {\left( {x-2} \right)^3}\;-x\left( {x + 1} \right)\left( {x-1} \right) + 6x\left( {x-3} \right) + 5x\\\,\,\,\,\,\,\,\begin{array}{*{20}{l}}{ = {x^3}\;-3.{x^2}.2 + 3x{{.2}^2}\;-{2^3}\;-x\left( {{x^2}\;-1} \right) + 6{x^2}\;-18x + 5x}\\\begin{array}{l} = {x^3}\;-6{x^2}\; + 12x-8-{x^3}\; + x + 6{x^2}\;-18x + 5x\\ =  – 8\end{array}\end{array}\\ \Rightarrow P = Q\end{array}\)

\(\begin{array}{l}{(a – b)^3} = {a^3} – 3{a^2}b + 3a{b^2} – {b^3} = {a^3} – {b^3} – 3ab(a – b)\\ \Rightarrow {a^3} – {b^3} = {(a – b)^3} + 3ab(a – b)\\ \Leftrightarrow Q = {(a – b)^3} + 3ab(a – b)\end{array}\)

Thay \(a + b = 5\) và \(ab =  – 3\) vào Q ta có

\(\begin{array}{c}Q = {(a – b)^3} + 3ab(a – b)\\ = {4^3} + 3.( – 3).4\\ = 64 – 36\\ = 28\end{array}\)

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

Ta có:

\(\begin{array}{c}\;B = {a^3}\; + {b^3}\; + {c^3}\;-3abc\;\\ = {(a + b)^3} – 3ab(a + b) + {c^3} – 3abc\\ = {(a + b)^3} + {c^3} – 3ab(a + b + c)\end{array}\)

Tương tự, ta có \({(a + b + c)^3} – 3(a + b)c(a + b + c)\)

\( \Rightarrow B = {(a + b + c)^3} – 3(a + b)c(a + b + c) – 3ab(a + b + c)\)

Mà \(\;a + b + c = 0\) nên \(\;B = 0 – 3(a + b)c.0 – 3ab.0 = 0\)