Question

If $a,b,c \in R$ , $x = {a^2} - bc,y = {b^2} - ca,z = {c^2} - ab$ then prove that ${x^3} + {y^3} + {z^3} - 3xyz$ is a perfect square.

Answer 1

Hint: Decompose ${x^3} + {y^3} + {z^3} - 3xyz$ using the formula ${x^3} + {y^3} + {z^3} - 3xyz = \dfrac{1}{2}\left( {x + y + z} \right)\left\{ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right\}$ . And substitute the values of x, y, z given in the question whenever required to prove it as a perfect square.


Complete step-by-step solution:
We are given that $a,b,c \in R$ and $x = {a^2} - bc,y = {b^2} - ca,z = {c^2} - ab$ . We have to prove that ${x^3} + {y^3} + {z^3} - 3xyz$ is a perfect square.
${x^3} + {y^3} + {z^3} - 3xyz = \dfrac{1}{2}\left( {x + y + z} \right)\left\{ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right\}$
Substitute the values of x, y, z given in the question as $x = {a^2} - bc,y = {b^2} - ca,z = {c^2} - ab$ in $x + y + z$
$
  x + y + z = {a^2} - bc + {b^2} - ca + {c^2} - ab \\
   = \dfrac{1}{2} \times 2\left( {{a^2} - bc + {b^2} - ca + {c^2} - ab} \right) \\
   = \dfrac{1}{2}\left( {2{a^2} - 2bc + 2{b^2} - 2ca + 2{c^2} - 2ab} \right) \\
   = \dfrac{1}{2}\left( {{a^2} - 2ab + {b^2} + {b^2} - 2bc + {c^2} + {a^2} - 2ac + {c^2}} \right) \\
   = \dfrac{1}{2}\left( {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right) \\
  \left( {\because {{\left( {a - b} \right)}^2} = {a^2} - 2ab + {b^2}} \right) \\
 $
Substitute the values of x, y, z given in the question as $x = {a^2} - bc,y = {b^2} - ca,z = {c^2} - ab$ in ${\left( {x - y} \right)^2}$
$
  {\left( {x - y} \right)^2} = {\left( {{a^2} - bc - \left( {{b^2} - ca} \right)} \right)^2} \\
   = {\left( {{a^2} - bc - {b^2} + ca} \right)^2} \\
   = {\left( {\left( {{a^2} - {b^2}} \right) + c\left( {a - b} \right)} \right)^2} \\
   = {\left( {\left( {a + b} \right)\left( {a - b} \right) + c\left( {a - b} \right)} \right)^2} \\
   = {\left( {\left( {a + b + c} \right)\left( {a - b} \right)} \right)^2} \\
   = {\left( {a + b + c} \right)^2}{\left( {a - b} \right)^2} \\
 $
Substitute the values of x, y, z given in the question as $x = {a^2} - bc,y = {b^2} - ca,z = {c^2} - ab$ in ${\left( {y - z} \right)^2}$
$
  {\left( {y - z} \right)^2} = {\left( {{b^2} - ca - \left( {{c^2} - ab} \right)} \right)^2} \\
   = {\left( {{b^2} - ca - {c^2} + ab} \right)^2} \\
   = {\left( {\left( {{b^2} - {c^2}} \right) + a\left( {b - c} \right)} \right)^2} \\
   = {\left( {\left( {b + c} \right)\left( {b - c} \right) + a\left( {b - c} \right)} \right)^2} \\
   = {\left( {\left( {a + b + c} \right)\left( {b - c} \right)} \right)^2} \\
   = {\left( {a + b + c} \right)^2}{\left( {b - c} \right)^2} \\
 $
Substitute the values of x, y, z given in the question as $x = {a^2} - bc,y = {b^2} - ca,z = {c^2} - ab$ in ${\left( {z - x} \right)^2}$
$
  {\left( {z - x} \right)^2} = {\left( {{c^2} - ab - \left( {{a^2} - bc} \right)} \right)^2} \\
   = {\left( {{c^2} - ab - {a^2} + bc} \right)^2} \\
   = {\left( {\left( {{c^2} - {a^2}} \right) + b\left( {c - a} \right)} \right)^2} \\
   = {\left( {\left( {c + a} \right)\left( {c - a} \right) + b\left( {c - a} \right)} \right)^2} \\
   = {\left( {\left( {a + b + c} \right)\left( {c - a} \right)} \right)^2} \\
   = {\left( {a + b + c} \right)^2}{\left( {c - a} \right)^2} \\
 $
Now substitute the obtained results in ${x^3} + {y^3} + {z^3} - 3xyz = \dfrac{1}{2}\left( {x + y + z} \right)\left\{ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right\}$
The value of $x + y + z = \dfrac{1}{2}\left( {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right)$
$
  {x^3} + {y^3} + {z^3} - 3xyz = \dfrac{1}{2}\left( {x + y + z} \right)\left\{ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right\} \\
  x + y + z = \dfrac{1}{2} \times \dfrac{1}{2}\left\{ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right\}\left\{ {{{\left( {a + b + c} \right)}^2}{{\left( {a - b} \right)}^2} + {{\left( {a + b + c} \right)}^2}{{\left( {b - c} \right)}^2} + {{\left( {a + b + c} \right)}^2}{{\left( {c - a} \right)}^2}} \right\} \\
 $
Take out (a+b+c) common from the second term.
 $
   = \dfrac{1}{4} \times \left\{ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right\}\left\{ {{{\left( {a + b + c} \right)}^2}\left( {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right)} \right\} \\
   = {\left( {\dfrac{{a + b + c}}{2}} \right)^2}{\left\{ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right\}^2} \\
   = {\left[ {\left( {\dfrac{{a + b + c}}{2}} \right)\left\{ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right\}} \right]^2} \\
 $
Therefore, ${x^3} + {y^3} + {z^3} - 3xyz$ can be written as ${\left[ {\left( {\dfrac{{a + b + c}}{2}} \right)\left\{ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right\}} \right]^2}$ which is a perfect square.


Note: In mathematics, a square number or a perfect square is an integer that is the square of another integer; in other words, it is the product of some integer with itself. For example, 16 is a square number, since it can be written as 4×4. Square numbers are non-negative. The square of a number is always positive