Question

Find the factors of \[{{\left( bc+ca+ab \right)}^{3}}-{{b}^{3}}{{c}^{3}}-{{c}^{3}}{{a}^{3}}-{{a}^{3}}{{b}^{3}}\] .

Answer 1

Hint: First of all assume \[bc=X\] , \[ca=Y\] , and \[ab=Z\] . Now, our new expression will be \[{{\left( X+Y+Z \right)}^{3}}-{{X}^{3}}-{{Y}^{3}}-{{Z}^{3}}\] . Transform the given equation as \[\left[ {{\left( X+Y+Z \right)}^{3}}-{{X}^{3}} \right]-\left[ \left( {{Y}^{3}}+{{Z}^{3}} \right) \right]\] . Now, use the identities \[\left( {{a}^{3}}-{{b}^{3}} \right)=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)\] and \[\left( {{a}^{3}}+{{b}^{3}} \right)=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)\] to simplify the expression \[\left[ {{\left( X+Y+Z \right)}^{3}}-{{X}^{3}} \right]-\left[ \left( {{Y}^{3}}+{{Z}^{3}} \right) \right]\] . Then, use the identity, \[{{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac\] and solve it further. Finally, replace \[bc\] by X, \[ca\] by Y, and \[ab\] by Z. Now, conclude the factors.

Complete step-by-step answer:
According to the question, it is given that we have to factorize,
\[{{\left( bc+ca+ab \right)}^{3}}-{{b}^{3}}{{c}^{3}}-{{c}^{3}}{{a}^{3}}-{{a}^{3}}{{b}^{3}}\] ………………..(1)
Let us assume,
\[bc=X\] …………………(2)
\[ca=Y\] …………………..(3)
\[ab=Z\] …………………..(4)
Using equation (2), equation (3), and equation (4), we can transform equation (1).
Transforming equation (1), we get
\[{{\left( bc+ca+ab \right)}^{3}}-{{b}^{3}}{{c}^{3}}-{{c}^{3}}{{a}^{3}}-{{a}^{3}}{{b}^{3}}\]
\[={{\left( X+Y+Z \right)}^{3}}-{{X}^{3}}-{{Y}^{3}}-{{Z}^{3}}\] ………………(5)
We know the identity, \[\left( {{a}^{3}}-{{b}^{3}} \right)=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)\] ……………………(4)
Replacing a by (X+Y+Z) and b by X in equation (4), we get
\[\left[ {{\left( X+Y+Z \right)}^{3}}-{{X}^{3}} \right]-\left[ \left( {{Y}^{3}}+{{Z}^{3}} \right) \right]\]
\[=\left\{ \left( X+Y+Z \right)-X \right\}\left\{ {{\left( X+Y+Z \right)}^{2}}+{{X}^{2}}+X\left( X+Y+Z \right) \right\}\] ……………….(5)
We also know the identity, \[\left( {{a}^{3}}+{{b}^{3}} \right)=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)\] ……………………(6)
Replacing a by Y and b by Z in equation (6), we get
\[\left( {{Y}^{3}}+{{Z}^{3}} \right)\]
\[=\left( Y+Z \right)\left( {{Y}^{2}}+{{Z}^{2}}-YZ \right)\] ……………………(7)
Now, arranging equation (5), we get
\[={{\left( X+Y+Z \right)}^{3}}-{{X}^{3}}-{{Y}^{3}}-{{Z}^{3}}\]
\[=\left[ {{\left( X+Y+Z \right)}^{3}}-{{X}^{3}} \right]-\left[ \left( {{Y}^{3}}+{{Z}^{3}} \right) \right]\] ……………………….(8)
Now, using equation (5) and equation (7) in equation (8), we get
\[=\left[ \left\{ \left( X+Y+Z \right)-X \right\}\left\{ {{\left( X+Y+Z \right)}^{2}}+{{X}^{2}}+X\left( X+Y+Z \right) \right\} \right]-\left[ \left( Y+Z \right)\left( {{Y}^{2}}+{{Z}^{2}}-YZ \right) \right]\] ……………(9)
We know the identity, \[{{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac\] …………………..(10)
Replacing a by X, b by Y, and c by Z in equation (10), we get
\[{{\left( X+Y+Z \right)}^{2}}=\left( {{X}^{2}}+{{Y}^{2}}+{{Z}^{2}}+2XY+2YZ+2XZ \right)\] ……………(11)
From equation (9) and equation (11), we have
\[\begin{align}
  & =\left[ \left\{ \left( X+Y+Z \right)-X \right\}\left\{ {{\left( X+Y+Z \right)}^{2}}+{{X}^{2}}+X\left( X+Y+Z \right) \right\} \right]-\left[ \left( Y+Z \right)\left( {{Y}^{2}}+{{Z}^{2}}-YZ \right) \right] \\
 & =\left[ \left( Y+Z \right)\left\{ {{X}^{2}}+{{Y}^{2}}+{{Z}^{2}}+2XY+2YZ+2XZ+{{X}^{2}}+{{X}^{2}}+XY+XZ \right\} \right]-\left[ \left( Y+Z \right)\left( {{Y}^{2}}+{{Z}^{2}}-YZ \right) \right] \\
 & =\left( Y+Z \right)\left[ {{Y}^{2}}+{{Z}^{2}}+{{X}^{2}}+{{X}^{2}}+{{X}^{2}}+2XY+XY+2XZ+XZ+2YZ \right]-\left( Y+Z \right)\left[ \left( {{Y}^{2}}+{{Z}^{2}}-YZ \right) \right] \\
\end{align}\]
Taking (Y+Z) as common in whole, we get
\[\begin{align}
  & =\left( Y+Z \right)\left[ \left( {{Y}^{2}}+{{Z}^{2}}+3{{X}^{2}}+3XY+3XZ+2YZ \right)-\left( {{Y}^{2}}+{{Z}^{2}}-YZ \right) \right] \\
 & =\left( Y+Z \right)\left[ \left( {{Y}^{2}}+{{Z}^{2}}+3{{X}^{2}}+3XY+3XZ+2YZ-{{Y}^{2}}-{{Z}^{2}}+YZ \right) \right] \\
 & =\left( Y+Z \right)\left[ \left( 3{{X}^{2}}+3XY+3XZ+3YZ \right) \right] \\
 & =\left( Y+Z \right)\left[ 3X\left( X+Y \right)+3Z\left( X+Y \right) \right] \\
 & =\left( Y+Z \right)\left[ \left( 3X+3Z \right)\left( X+Y \right) \right] \\
 & =\left( Y+Z \right)3\left( X+Z \right)\left( X+Y \right) \\
\end{align}\]
\[=3\left( Y+Z \right)\left( X+Z \right)\left( X+Y \right)\] ……………………(12)
From equation (2), equation (3), and equation (4), we have \[bc=X\] , \[ca=Y\] , and \[ab=Z\] .
Now, putting the value of X, Y, and Z in the equation (12), we get
\[\begin{align}
  & 3\left( Y+Z \right)\left( X+Z \right)\left( X+Y \right) \\
 & =3\left( ca+ab \right)\left( bc+ab \right)\left( bc+ca \right) \\
\end{align}\]
Hence, the factors of \[{{\left( bc+ca+ab \right)}^{3}}-{{b}^{3}}{{c}^{3}}-{{c}^{3}}{{a}^{3}}-{{a}^{3}}{{b}^{3}}\] is \[3,\left( ca+ab \right),\left( bc+ab \right),\text{and}\,\left( bc+ca \right)\] .

Note: To solve this question one can think to expand the expression \[{{\left( bc+ca+ab \right)}^{3}}-{{b}^{3}}{{c}^{3}}-{{c}^{3}}{{a}^{3}}-{{a}^{3}}{{b}^{3}}\] using the identity, \[{{\left( a+b+c \right)}^{3}}={{a}^{3}}+{{b}^{3}}+{{c}^{3}}+3{{a}^{2}}\left( b+c \right)+3{{b}^{2}}\left( a+c \right)+3{{c}^{2}}\left( a+b \right)+6abc\] . When we use this identity, we are only expanding the expression. But we have to find its factors, so only expanding it will not work here.