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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...

If a,b,c∈Ra,b,c \in R , x=a2−bc,y=b2−ca,z=c2−abx = {a^2} - bc,y = {b^2} - ca,z = {c^2} - ab then prove that x3+y3+z3−3xyz{x^3} + {y^3} + {z^3} - 3xyz is a perfect square.

Explanation

Solution

Decompose x3+y3+z3−3xyz{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∈Ra,b,c \in R and x=a2−bc,y=b2−ca,z=c2−abx = {a^2} - bc,y = {b^2} - ca,z = {c^2} - ab . We have to prove that x3+y3+z3−3xyz{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=a2−bc,y=b2−ca,z=c2−abx = {a^2} - bc,y = {b^2} - ca,z = {c^2} - ab in x+y+zx + y + z
x+y+z=a2−bc+b2−ca+c2−ab =12×2(a2−bc+b2−ca+c2−ab) =12(2a2−2bc+2b2−2ca+2c2−2ab) =12(a2−2ab+b2+b2−2bc+c2+a2−2ac+c2) =12((a−b)2+(b−c)2+(c−a)2) (∵(a−b)2=a2−2ab+b2)  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=a2−bc,y=b2−ca,z=c2−abx = {a^2} - bc,y = {b^2} - ca,z = {c^2} - ab in (x−y)2{\left( {x - y} \right)^2}
(x−y)2=(a2−bc−(b2−ca))2 =(a2−bc−b2+ca)2 =((a2−b2)+c(a−b))2 =((a+b)(a−b)+c(a−b))2 =((a+b+c)(a−b))2 =(a+b+c)2(a−b)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=a2−bc,y=b2−ca,z=c2−abx = {a^2} - bc,y = {b^2} - ca,z = {c^2} - ab in (y−z)2{\left( {y - z} \right)^2}
(y−z)2=(b2−ca−(c2−ab))2 =(b2−ca−c2+ab)2 =((b2−c2)+a(b−c))2 =((b+c)(b−c)+a(b−c))2 =((a+b+c)(b−c))2 =(a+b+c)2(b−c)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=a2−bc,y=b2−ca,z=c2−abx = {a^2} - bc,y = {b^2} - ca,z = {c^2} - ab in (z−x)2{\left( {z - x} \right)^2}
(z−x)2=(c2−ab−(a2−bc))2 =(c2−ab−a2+bc)2 =((c2−a2)+b(c−a))2 =((c+a)(c−a)+b(c−a))2 =((a+b+c)(c−a))2 =(a+b+c)2(c−a)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=12((a−b)2+(b−c)2+(c−a)2)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, x3+y3+z3−3xyz{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