Evaluating Symmetrical Expressions

Algebra Level 4

( x 3 − y 3 ) 2 ( x y ) 3 + ( y 3 − z 3 ) 2 ( y z ) 3 + ( z 3 − x 3 ) 2 ( z x ) 3 \dfrac{ ( x^3 - y^3) ^2 } { (xy)^3 } + \dfrac{ ( y^3 - z^3 )^2 } { (yz)^3} + \dfrac{ ( z^3 - x^3 ) ^2 } { ( zx)^3 } If x 3 + y 3 + z 3 = 0 x^3 + y^3 + z^3 = 0 and x y z ≠ 0 xyz \neq 0 , what is the value of the expression above.


The answer is -9.

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Calvin Lin by Calvin Lin

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7 solutions

Rui-Xian Siew
Aug 23, 2016

Starting with given condition, ( x 3 + y 3 + z 3 ) ( x 6 + y 6 + z 6 ) = 0 ⇒ x 9 + y 9 + z 9 + x 3 ( y 3 + z 3 ) + y 3 ( x 3 + z 3 ) + z 3 ( y 3 + x 3 ) = 0 ⇒ x 9 + y 9 + z 9 = − [ x 3 ( y 3 + z 3 ) + y 3 ( x 3 + z 3 ) + z 3 ( y 3 + x 3 ) ] L e t x 9 + y 9 + z 9 ( x y z ) 3 = A S = ( x 3 − y 3 ) 2 ( x y ) 3 + ( x 3 − z 3 ) 2 ( x z ) 3 + ( z 3 − y 3 ) 2 ( z y ) 3 S = ( x 6 + y 6 ) − 2 ( x y ) 3 ( x y ) 3 + ( x 6 + z 6 ) − 2 ( x z ) 3 ( x z ) 3 + ( z 6 + y 6 ) − 2 ( z y ) 3 ( z y ) 3 S = z 3 ( x 6 + y 6 ) + x 3 ( z 6 + y 6 ) + y 3 ( x 6 + z 6 ) ( x y z ) 3 − 6 ⇒ S = − A − 6 S = z 3 ( x 6 + y 6 + 2 ( x y ) 3 − 2 ( x y ) 3 ) + x 3 ( z 6 + y 6 + 2 ( z y ) 3 − 2 ( z y ) 3 ) + y 3 ( x 6 + z 6 + 2 ( x z ) 3 − 2 ( x z ) 3 ) ( x y z ) 3 − 6 S = z 6 ( z 3 ) + x 6 ( x 3 ) + y 6 ( y 3 ) ( x y z ) 3 − 12 ⇒ S = A − 12 A − 12 = − A − 6 ⇒ A = 3 ∴ S = 3 − 12 = − 9 ({ x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 })({ x }^{ 6 }+{ y }^{ 6 }+{ z }^{ 6 })=0\\ \Rightarrow { x }^{ 9 }+{ y }^{ 9 }+{ z }^{ 9 }+{ x }^{ 3 }({ y }^{ 3 }+{ z }^{ 3 })+{ y }^{ 3 }({ x }^{ 3 }+{ z }^{ 3 })+{ z }^{ 3 }({ y }^{ 3 }+{ x }^{ 3 })=0\\ \Rightarrow { x }^{ 9 }+{ y }^{ 9 }+{ z }^{ 9 }=-\left[ { x }^{ 3 }({ y }^{ 3 }+{ z }^{ 3 })+{ y }^{ 3 }({ x }^{ 3 }+{ z }^{ 3 })+{ z }^{ 3 }({ y }^{ 3 }+{ x }^{ 3 }) \right] \\ Let\quad \frac { { x }^{ 9 }+{ y }^{ 9 }+{ z }^{ 9 } }{ { (xyz) }^{ 3 } } =A\\ S=\frac { { ({ x }^{ 3 }-{ y }^{ 3 }) }^{ 2 } }{ { (xy) }^{ 3 } } +\frac { { ({ x }^{ 3 }-{ z }^{ 3 }) }^{ 2 } }{ { (xz) }^{ 3 } } +\frac { { ({ z }^{ 3 }-{ y }^{ 3 }) }^{ 2 } }{ { (zy) }^{ 3 } } \\ S=\frac { { ({ x }^{ 6 }+{ y }^{ 6 }) }-2{ (xy) }^{ 3 } }{ { (xy) }^{ 3 } } +\frac { { ({ x }^{ 6 }+{ z }^{ 6 }) }-2{ (xz) }^{ 3 } }{ { (xz) }^{ 3 } } +\frac { { ({ z }^{ 6 }+{ y }^{ 6 }) }-2{ (zy) }^{ 3 } }{ { (zy) }^{ 3 } } \\ S=\frac { { { z }^{ 3 }({ x }^{ 6 }+{ y }^{ 6 }) }+x^{ 3 }({ z }^{ 6 }+{ y }^{ 6 })+{ y }^{ 3 }({ x }^{ 6 }+z^{ 6 }) }{ { (xyz) }^{ 3 } } -6\quad \Rightarrow S=-A-6\\ S=\frac { { { z }^{ 3 }({ x }^{ 6 }+{ y }^{ 6 }+2{ (xy) }^{ 3 }-2{ (xy) }^{ 3 }) }+x^{ 3 }({ z }^{ 6 }+{ y }^{ 6 }+2{ (zy) }^{ 3 }-2{ (zy) }^{ 3 })+{ y }^{ 3 }({ x }^{ 6 }+z^{ 6 }+2{ (xz) }^{ 3 }-2{ (xz) }^{ 3 }) }{ { (xyz) }^{ 3 } } -6\\ S=\frac { { { z }^{ 6 }({ z }^{ 3 }) }+x^{ 6 }({ x }^{ 3 })+{ y }^{ 6 }({ y }^{ 3 }) }{ { (xyz) }^{ 3 } } -12\quad \Rightarrow S=A-12\\ A-12=-A-6\quad \Rightarrow A=3\\ \therefore \quad S=3-12=-9

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Moderator note:

Nicely done. Can you generalize the problem, replacing 3 with n n (but leaving the squares as they are)?

If n is odd integer, given x n + y n + z n = 0 a n d x y z ≠ 0 ( x n + y n + z n ) ( x 2 n + y 2 n + z 2 n ) = 0 ⇒ x 3 n + y 3 n + z 3 n + x n ( y n + z n ) + y n ( x n + z n ) + z n ( y n + x n ) = 0 ⇒ x 3 n + y 3 n + z 3 n = − [ x n ( y n + z n ) + y n ( x n + z n ) + z n ( y n + x n ) ] L e t x 3 n + y 3 n + z 3 n ( x y z ) n = A S = ( x n − y n ) 2 ( x y ) n + ( x n − z n ) 2 ( x z ) n + ( z n − y n ) 2 ( z y ) n S = ( x 2 n + y 2 n ) − 2 ( x y ) n ( x y ) n + ( x 2 n + z 2 n ) − 2 ( x z ) n ( x z ) n + ( z 2 n + y 2 n ) − 2 ( z y ) n ( z y ) n S = z n ( x 2 n + y 2 n ) + x n ( z 2 n + y 2 n ) + y n ( x 2 n + z 2 n ) ( x y z ) n − 6 ⇒ S = − A − 6 S = z n ( x 2 n + y 2 n + 2 ( x y ) n − 2 ( x y ) n ) + x n ( z 2 n + y 2 n + 2 ( z y ) n − 2 ( z y ) n ) + y n ( x 2 n + z 2 n + 2 ( x z ) n − 2 ( x z ) n ) ( x y z ) n − 6 S = z n ( z 2 n ) + x n ( x 2 n ) + y n ( y 2 n ) ( x y z ) n − 12 ⇒ S = A − 12 2 S = − 18 ⇒ S = − 9 ∴ ( x n − y n ) 2 ( x y ) n + ( x n − z n ) 2 ( x z ) n + ( z n − y n ) 2 ( z y ) n = − 9 { x }^{ n }+{ y }^{ n }+{ z }^{ n }=0\quad and\quad xyz\neq 0\quad \\ ({ x }^{ n }+{ y }^{ n }+{ z }^{ n })({ x }^{ 2n }+{ y }^{ 2n }+{ z }^{ 2n })=0\\ \Rightarrow { x }^{ 3n }+{ y }^{ 3n }+{ z }^{ 3n }+{ x }^{ n }({ y }^{ n }+{ z }^{ n })+{ y }^{ n }({ x }^{ n }+{ z }^{ n })+{ z }^{ n }({ y }^{ n }+{ x }^{ n })=0\\ \Rightarrow { x }^{ 3n }+{ y }^{ 3n }+{ z }^{ 3n }=-\left[ { x }^{ n }({ y }^{ n }+{ z }^{ n })+{ y }^{ n }({ x }^{ n }+{ z }^{ n })+{ z }^{ n }({ y }^{ n }+{ x }^{ n }) \right] \\ Let\quad \frac { { x }^{ 3n }+{ y }^{ 3n }+{ z }^{ 3n } }{ { (xyz) }^{ n } } =A\\ S=\frac { { ({ x }^{ n }-{ y }^{ n }) }^{ 2 } }{ { (xy) }^{ n } } +\frac { { ({ x }^{ n }-{ z }^{ n }) }^{ 2 } }{ { (xz) }^{ n } } +\frac { { ({ z }^{ n }-{ y }^{ n }) }^{ 2 } }{ { (zy) }^{ n } } \\ S=\frac { { ({ x }^{ 2n }+{ y }^{ 2n }) }-2{ (xy) }^{ n } }{ { (xy) }^{ n } } +\frac { { ({ x }^{ 2n }+{ z }^{ 2n }) }-2{ (xz) }^{ n } }{ { (xz) }^{ n } } +\frac { { ({ z }^{ 2n }+{ y }^{ 2n }) }-2{ (zy) }^{ n } }{ { (zy) }^{ n } } \\ S=\frac { { { z }^{ n }({ x }^{ 2n }+{ y }^{ 2n }) }+x^{ n }({ z }^{ 2n }+{ y }^{ 2n })+{ y }^{ n }({ x }^{ 2n }+z^{ 2n }) }{ { (xyz) }^{ n } } -6\quad \Rightarrow S=-A-6\\ S=\frac { { { z }^{ n }({ x }^{ 2n }+{ y }^{ 2n }+2{ (xy) }^{ n }-2{ (xy) }^{ n }) }+x^{ n }({ z }^{ 2n }+{ y }^{ 2n }+2{ (zy) }^{ n }-2{ (zy) }^{ n })+{ y }^{ n }({ x }^{ 2n }+z^{ 2n }+2{ (xz) }^{ n }-2{ (xz) }^{ n }) }{ { (xyz) }^{ n } } -6\\ S=\frac { { { z }^{ n }({ z }^{ 2n }) }+x^{ n }({ x }^{ 2n })+{ y }^{ n }({ y }^{ 2n }) }{ { (xyz) }^{ n } } -12\quad \Rightarrow S=A-12\\ 2S=-18\quad \Rightarrow S=-9\\ \therefore \quad \frac { { ({ x }^{ n }-{ y }^{ n }) }^{ 2 } }{ { (xy) }^{ n } } +\frac { { ({ x }^{ n }-{ z }^{ n }) }^{ 2 } }{ { (xz) }^{ n } } +\frac { { ({ z }^{ n }-{ y }^{ n }) }^{ 2 } }{ { (zy) }^{ n } } =-9

Rui-Xian Siew - 4 years, 9 months ago

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How about n n even?

Did you need to use the fact that n n is odd? If so, you should explain where / why.

Calvin Lin Staff - 4 years, 9 months ago

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In the real domain, n with even parity would imply that each of x,y and z equals zero. However in the complex domain, solutions do exist. Here is a brief generalization: ∑ x , y , z ( x n − y n ) 2 x n y n = ∑ ( x n + y n ) 2 − 4 x n y n x n y n = ∑ z 2 n x n y n − 4 = − 12 + 1 x n y n z n ∑ z 3 n = − 12 + 3 = − 9 ( ∵ a + b + c = 0 ⇒ a 3 + b 3 + c 3 = 3 a b c , p u t t i n g a = x n , b = y n , c = z n ) \sum _{ x,y,z }^{ }{ \frac { { ({ x }^{ n }-{ y }^{ n }) }^{ 2 } }{ { x }^{ n }{ y }^{ n } } } =\quad \sum \frac { { ({ x }^{ n }+{ y }^{ n }) }^{ 2 }-4{ x }^{ n }{ y }^{ n } }{ { x }^{ n}{ y }^{ n } } =\quad \sum { \frac { { z }^{ 2n } }{ { x }^{ n }{ y }^{ n } } } -4\quad =\quad -12+\quad \frac { 1 }{ { x }^{ n }{ y }^{ n }{ z }^{ n } } \sum { { z }^{ 3n } } =\quad -12+3\quad =-9\quad \\ (\quad \because \quad a+b+c=0\quad \Rightarrow { a }^{ 3 }{ +b }^{ 3 }+{ c }^{ 3 }=3abc,\quad putting\quad a={ x }^{ n },b={ y }^{ n },c={ z }^{ n })

Aditya Dhawan - 4 years, 9 months ago

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@Aditya Dhawan – There was no restriction that the numbers had to be real.

Yes. That approach is what I envisioned, and is a clearer presentation of Rui's solution. Can you add that as a separate​ solution? Thanks!

Calvin Lin Staff - 4 years, 9 months ago

x 3 + z 3 = − y 3 x 3 + y 3 = − z 3 y 3 + z 3 = − x 3 x^{3}+z^{3}=-y^{3} \\ x^{3}+y^{3}=-z^{3} \\ y^{3}+z^{3}=-x^{3}

( x 3 − y 3 ) 2 ( x y ) 3 + ( y 3 − z 3 ) 2 ( y z ) 3 + ( z 3 − x 3 ) 2 ( z x ) 3 \dfrac{ ( x^3 - y^3) ^2 } { (xy)^3 } + \dfrac{ ( y^3 - z^3 )^2 } { (yz)^3} + \dfrac{ ( z^3 - x^3 ) ^2 } { ( zx)^3 }

= z 3 ( x 6 + y 6 − 2 x 3 y 3 ) + x 3 ( y 6 + z 6 − 2 y 3 z 3 ) + y 3 ( x 6 + z 6 − 2 x 3 z 3 ) x 3 y 3 z 3 = z 3 x 6 + z 3 y 6 − 2 x 3 y 3 z 3 + x 3 y 6 + x 3 z 6 − 2 x 3 y 3 z 3 + y 3 x 6 + y 3 z 6 − 2 x 3 y 3 z 3 x 3 y 3 z 3 = x 3 y 6 + y 3 x 6 + x 3 z 6 + z 3 x 6 + y 3 z 6 + z 3 y 6 − 6 x 3 y 3 z 3 ) x 3 y 3 z 3 = x 3 y 3 ( x 3 + y 3 ) + x 3 z 3 ( x 3 + z 3 ) + y 3 z 3 ( y 3 + z 3 ) − 6 x 3 y 3 z 3 x 3 y 3 z 3 = x 3 y 3 ( − z 3 ) + x 3 z 3 ( − y 3 ) + y 3 z 3 ( − x 3 ) − 6 x 3 y 3 z 3 x 3 y 3 z 3 = − 9 x 3 y 3 z 3 x 3 y 3 z 3 = − 9 = \dfrac{ z^{3}(x^{6}+y^{6}-2x^{3}y^{3})+x^{3}(y^{6}+z^{6}-2y^{3}z^{3})+y^{3}(x^{6}+z^{6}-2x^{3}z^{3}) } { x^{3}y^{3}z^{3} } \\ = \dfrac{z^{3}x^{6}+z^{3}y^{6}-2x^{3}y^{3}z^{3}+x^{3}y^{6}+x^{3}z^{6}-2x^{3}y^{3}z^{3}+y^{3}x^{6}+y^{3}z^{6}-2x^{3}y^{3}z^{3}}{x^{3}y^{3}z^{3}} \\ = \dfrac{x^{3}y^{6}+y^{3}x^{6}+x^{3}z^{6}+z^{3}x^{6}+y^{3}z^{6}+z^{3}y^{6}-6x^{3}y^{3}z^{3})}{x^{3}y^{3}z^{3}} \\ = \dfrac{x^{3}y^{3}(x^{3}+y^{3})+x^{3}z^{3}(x^{3}+z^{3})+y^{3}z^{3}(y^{3}+z^{3})-6x^{3}y^{3}z^{3}}{x^{3}y^{3}z^{3}} \\ = \dfrac{x^{3}y^{3}(-z^{3})+x^{3}z^{3}(-y^{3})+y^{3}z^{3}(-x^{3})-6x^{3}y^{3}z^{3}}{x^{3}y^{3}z^{3}} \\ = \dfrac{-9x^{3}y^{3}z^{3}}{x^{3}y^{3}z^{3}} \\=-9

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Solved the same way too!

tom engelsman - 4 years, 9 months ago
Aditya Dhawan
Aug 24, 2016

Here is a brief generalization: ∑ x , y , z ( x n − y n ) 2 x n y n = ∑ ( x n + y n ) 2 − 4 x n y n x n y n = ∑ z 2 n x n y n − 4 = − 12 + 1 x n y n z n ∑ z 3 n = − 12 + 3 = − 9 ( ∵ a + b + c = 0 ⇒ a 3 + b 3 + c 3 = 3 a b c , p u t t i n g a = x n , b = y n , c = z n ) \sum _{ x,y,z }^{ }{ \frac { { ({ x }^{ n }-{ y }^{ n }) }^{ 2 } }{ { x }^{ n }{ y }^{ n } } } =\quad \sum \frac { { ({ x }^{ n }+{ y }^{ n }) }^{ 2 }-4{ x }^{ n }{ y }^{ n } }{ { x }^{ n}{ y }^{ n } } =\quad \sum { \frac { { z }^{ 2n } }{ { x }^{ n }{ y }^{ n } } } -4\quad =\quad -12+\quad \frac { 1 }{ { x }^{ n }{ y }^{ n }{ z }^{ n } } \sum { { z }^{ 3n } } =\quad -12+3\quad =-9\quad \\ (\quad \because \quad a+b+c=0\quad \Rightarrow { a }^{ 3 }{ +b }^{ 3 }+{ c }^{ 3 }=3abc,\quad putting\quad a={ x }^{ n },b={ y }^{ n },c={ z }^{ n })

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Kushal Bose
Aug 24, 2016

I think there is no real exists that satisfy the given condition.

I have used complex number.

I have considered x = 1 , y 3 = ω , z 3 = ω 2 x=1,y^3=\omega,z^3=\omega^2 .

After putting this I have found answer is 9 9

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x = − 1 , y = − 1 , z = 2 3 x = -1, y = -1, z = \sqrt[3]{2} is a real solution to the conditions.

You have shown it for a specific instance. For a rigorous solution, explain why it would hold true for all possible cases.

Calvin Lin Staff - 4 years, 9 months ago
Sparsh Setia
Nov 28, 2016

it was a very easy question

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For posting of solutions, please restrict it to actual solutions of the problem.

You can post this as a comment instead.

Calvin Lin Staff - 4 years, 6 months ago
Rishav Koirala
Sep 14, 2016

I have one that's a bit cleaner- Noting that the expressions are similar but for the variables interchanged, take the first one and simplify it- ( x 3 − y 3 ) 2 ( x y ) 3 = ( x 2 y − y 2 x ) 2 x y = x 4 y 2 − 2 x y + y 4 x 2 x y \large \frac{(x^3-y^3)^2}{(xy)^3} = \frac{(\frac{x^2}{y}-\frac{y^2}{x})^2}{xy} = \frac{\frac{x^4}{y^2} -2xy + \frac{y^4}{x^2}}{xy} , which can then be written as x 3 y 3 + y 3 x 3 − 2 \frac{x^3}{y^3} + \frac{y^3}{x^3} -2 Now, the remaining two expressions can also be written in a similar fashion, so the whole expression becomes x 3 y 3 + z 3 y 3 + y 3 x 3 + z 3 x 3 + x 3 z 3 + y 3 z 3 − 6 \frac{x^3}{y^3} + \frac{z^3}{y^3}+\frac{y^3}{x^3}+\frac{z^3}{x^3}+\frac{x^3}{z^3}+\frac{y^3}{z^3} -6 Now grouping together the terms with the same denominator and using the fact that x 3 + y 3 + z 3 = 0 x^3+y^3+z^3 =0 , the answer is − 1 − 1 − 1 − 6 = − 9 -1-1-1-6 = \boxed{-9}

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Rajen Kapur
Aug 26, 2016

At the outset the problem can be simplified to evaluate ( p − q ) 2 p q + ( q − r ) 2 q r + ( r − p ) 2 r p \frac{(p-q)^2}{pq}+\frac {(q-r)^2}{qr} +\frac {(r-p)^2}{rp} where p + q + r = 0 p+q+r=0 and p q r ≠ 0 pqr\neq 0 with definition of new variables as p = x 3 , q = y 3 , r = z 3 p=x^3,q=y^3,r=z^3 , knowing that x y z ≠ 0 ⟹ x 3 y 3 z 3 ≠ 0. xyz\neq 0 \implies x^3y^3z^3 \neq 0.

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