Complex solutions

Since the given conditions, we have: $(x+y+z)\big( (x-y)^2 + (y-z)^2 + (z-x)^2 \big) = 20xyz.$ Or equivalently, $(x+y+z)(x^2+y^2+z^2-xy-yz-zx) = 10xyz.$ On the other hand, we have $(x+y+z)(x^2+y^2+z^2-xy-yz-zx) = x^3+y^3+z^3-3xyz.$ Therefore, $x^3+y^3+z^3 = 13xyz$ and $\frac{ x^3 + y^3 + z^3}{xyz} = 13$ .

We know that $x^3+y^3+z^3-3xyz=\frac{1}{2}(x+y+z)((x-y)^2+(x-z)^2+(y-z)^2) = 10xyz$ . Dividing both sides by $xyz$ and adding $3$ to both sides gives $\frac{x^3+y^3+z^3}{xyz}=\boxed{13}$ .

If you didn't know the first fact, I recommend Titu Andreescu's "Mathematical Olympiad Treasures" to you. Great read.

$\frac{x^3 + y^3 + z ^3}{xyz} \\ = \frac{(x+y+z)(x^2 + y^2 + z^2 - xy - yz -zx) + 3xyz}{xyz} \\ = \frac{(20)(\frac{xyz}{2}) + 3xyz}{xyz} = 10 + 3 = 13$

We use the identity, $x^3+y^3+z^3-3xyz=\frac{1}{2}(x+y+z)((x-y)^2+(y-z)^2+(z-x)^2)$ .Then $\frac{x^3+y^3+z^3}{xyz}=\frac{1}{2}(x+y+z)+3=10+3=13$ .

$\frac{x^3+y^3+z^3}{xyz}$ $=\frac{(x+y+z)\{(x-y)^2+(y-z)^2+(z-x)^2\}}{2*(xyz)}+3$ Putting the appropriate values as given in the problem, we get, $\frac{x^3+y^3+z^3}{xyz}=\fbox{13}$

First note that : $(x+y+z)\left[(x-y)^2+(x-z)^2+(y-z)^2\right]=x^3+y^3+z^3-3 x y z.$

Multiply the second equation with $x+y+z$ , to get : $2(x^3+y^3+z^3-3x y z)=20x y z \Longrightarrow 2(x^3+y^3+z^3)=26x y z.$ From here we see that : $\frac{x^3+y^3+z^3}{xy z}= \frac{26}{2} = \boxed{13}.$

My solution using polynomials: Let $x+y+z=p,xy+zx+yz=q,xyz=r$ Equation 1 gives that $p=20$ Equation 2 gives on expansion: $2(x^2+y^2+z^2)-2(xy+yz+zx)=xyz \Rightarrow x^2+y^2+z^2=\frac{r+2q}{2}$

Let $P(m)$ be a polynomial such that $P(m)=(m-x)(m-y)(m-z)=m^3-p m^2+q m-r$ so $P(x)=P(y)=P(z)=0$ .

$P(x)=0 \Rightarrow x^3-px^2+qx-r=0 \Rightarrow x^3=px^2-qx+r$ and similarly for $y,z$

Adding the three equations together gives: $x^3+y^3+z^3=p(x^2+y^2+z^2)-(x+y+z)+3r=p(\frac{r+2q}{2})-pq+3r=\frac{pr+6r}{2}$ by the earlier equation.

Hence we have: $\frac{x^3+y^3+z^3}{xyz}=\frac{(\frac{pr+6r}{2})}{r}=\frac{p+6}{2}=\frac{20+6}{2}=13$

Observe that $\begin{aligned} (x+y+z)^3 &= \sum (x^3) + 3\sum(x^2(y+z)) + 6xyz \\ &= -2\sum(x^3) + 3\sum( x^2( x+ y+ z) ) + 6xyz \text{; Add and subtract } 3\sum(x^3) \\ &= -2\sum(x^3) + 3(x+y+z) \sum( x^2 ) + 6xyz \\ \end{aligned} \\$

Now, squaring the first equation and adding it to the second equation (given ones) gives us: $3\sum( x^2 ) = 20^2 + xyz$

Plugging this in the result we obtained gives: $\begin{aligned} (x+y+z)^3 &= -2\sum(x^3) + (x+y+z)(20^2 + xyz) + 6xyz \end{aligned}$

Simplifying and plugging the value of $x+y+z = 20$ gives the answer as $13$ .

Man! I had to think a heck lot for this :D

Label the two given equations $(1)$ and $(2).$ Expanding $(2$ ) gives $\qquad 2(x^2+y^2+z^2)- 2(xy+xz+yz) = xyz. \qquad (3)$ Now we square both sides of $(1)$ to get $x^2+y^2+z^2+2(xy+xz+yz) = 400.$ We multiply this equation by $2$ and subtract equation $(3)$ : $\qquad 6(xy+xz+yz) = 800-xyz. \qquad (4)$

At this point we return to equation $(1)$ , which we cube both sides of and factor: $x^3+y^3+z^3+3[xy(x+y)+xz(x+z)+yz(y+z)]$ $+6xyz = 8000.$ Using equation $(1),$ we can replace each of the sums $x+y,x+z,y+z$ with $20-z,20-y,20-x.$ Therefore, we have $x^3+y^3+z^3+3[xy(20-z)+xz(20-y)+yz(20-x)]$ $+ 6xyz = 8000$ $x^3+y^3+z^3+60(xy+xz+yz) - 3xyz = 8000$

We multiply equation $(4)$ by $10$ to get $60(xy+xz+yz) = 8000 - 10xyz.$ Now we substitute into the previous equation: $x^3+y^3+z^3+8000-10xyz-3xyz=8000,$ or $x^3+y^3+z^3 = 13xyz.$ Thus, $\dfrac{x^3+y^3+z^3}{xyz} = \boxed{13}.$

we know that

$x^{3}+y^{3}+z^{3}=(x+y+z)^{3}-6xyz-3xyz(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})+9xyz$

From two given equation we have

$\frac{(x+y+z)^{3}}{3xyz}-\frac{1}{6}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$

hence

$\frac{x^{3}+y^{3}+z^{3}}{xyz}=\frac{20^{3}}{xyz}-6-(\frac{20^{3}}{xyz}-10)+9$

$\frac{x^{3}+y^{3}+z^{3}}{xyz}=\boxed{13}$

We know that x^3+y^3+z^3-3xyz=(0.5)(x+y+z){(x−y)^2+(y−z)^2+(z−x)^2}
=>x^3+y^3+z^3-3xyz=10xyz(putting the values)
=>x^3+y^3+z^3=13xyz
=>(x^3+y^3+z^3)/xyz=13 . (sorry , i don't know how to use latex )

We have $x^3+y^3+z^3-3xyz= \frac 12 (x+y+z) \left[(x-y)^2+(y-z)^2+(z-x)^2 \right]$ . Thus, $x^3+y^3+z^3=13xyz$

$\sin a + \sin b = 2 \sin (\frac {a+b}{2})\cos (\frac {a-b }{2})$

Similarly for $c$ and $d$ . Adding both gives:

$0=\sin a+\sin b+\sin c+\sin d\ =$

(using $c+d=-(a+b)$ )

$=2\sin (\frac {a+b}{2})(\cos (\frac {a-b}{2})-\cos (\frac {c-d}{2})) =$

(using $c-d=(a+c) + (b+c)$ and $a-b=(a+c) - (b+c)$ )

$= 4\sin (\frac {a+b}{2})\sin (\frac {a+c}{2})\sin (\frac {b+c}{2})$

which leads to: $a=-b$ or $a=-c$ or $b=-c$ .

Now use the counting already explained by Thomas B.

Let a cubic equation of the form $n^{3} + an^{2} + bn + c = 0$ have roots equivalent to x, y, and z.

By Vieta's theorem, the root-coefficient relationships are written as follows: x+y+z = -a, xy + yz + xz = b, xyz = -c.

The first equation in the problem can be combined with the first root-coefficient relationship to yield a = -20. The second equation in the problem can be expanded and simplified as follows: $2(x^{2}+y^{2}+z^{2})-2(xy+yz+xz)=xyz$ $2((x+y+z)^{2}-2(xy+yz+xz))-2(xy+yz+xz)=xyz$ $2(x+y+z)^{2}-6(xy+yz+xz)=xyz$

The root-coefficient relationships then allow this to be rewritten as: $2a^{2}-6b=-c$ , or $c=-2(a^{2}-3b)$

The final expression, $\frac{x^{3}+y^{3}+z^{3}}{xyz}$ , is simplified in this manner: = $\frac{x^{3}+y^{3}+z^{3}-3xyz+3xyz}{xyz}$ = $\frac{(x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-xz)+3xyz}{xyz}$ = $\frac{(x+y+z)((x+y+z)^{2}-3(xy-yz-xz))+3xyz}{xyz}$ Writing in terms of a, b, and c: = $\frac{(-a)(a^{2}-3b)-3c}{-c}$ = $3+\frac{(a)(a^{2}-3b)}{c}$

From the earlier calculations: $c=-2(a^{2}-3b)$ , a = -20: = $3+\frac{(a)(a^{2}-3b)}{-2(a^{2}-3b}$ = $3+\frac{a}{-2}$ = $3+\frac{-20}{-2}$

=13

We have $x^3+y^3+z^3-3xyz=(x+y)^3-3xy(x+y)+z^3-3xyz$

$=(x+y+z)((x+y)^2-(x+y)z+z^2) -3xy(x+y+z)=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$ $=(x+y+z)(\frac{1}{2}((x-y)^2+(y-z)^2+(z-x)^2))$

Now, by substituting the values given, we have: $x^3+y^3+z^3-3xyz=20\frac{1}{2}(xyz)$ or $x^3+y^3+z^3=13xyz$

Hence, 13 is the answer we desired.

Note that $(x + y + z)[(x - y)^2 + (y - z)^2 + (z - x)^2] = 2(x^3 + y^3 + z^3 - 3xyz)$ . So,

$26xyz = 2(x^3 + y^3 + z^3)$ and thus $\frac{x^3 + y^3 + z^3}{xyz} = 13$ .

We will need 3 equations to solve for the values of all three numbers, since we were given only $2$ we need to express the desired expression in terms of the variables above.

Expanding the LHS of the second equation will yield,

$2(x^{2} + y^{2} + z^{2} - xy - yz - xz) = xyz$

Multiplying the LHS of the first equation to both sides of the second equation will yield:

$2x^{3} + 2y^{3} + 2z^{3} - 6xyz = xyz (x + y + z)$

Now if we solve for $x^{3} + y^{3} + z^{3}$ in the new equation, we shall have the numerator of the algebraic expression we are asked to find. So,

$x^{3} + y^{3} + z^{3} = (1/2)(xyz(x + y + z) + 6xyz)$

With then if we divide it by $xyz$ we'll arrive at the desired answer. Thus,

$\frac{x^{3} + y^{3} + z^{3}}{xyz} = (1/2)(x + y + z + 6)$

And substituting $20$ to $x + y + z$ we will get the answer:

$\frac{x^{3} + y^{3} + z^{3}}{xyz} = (1/2)(20 + 6)$

$\frac{x^{3} + y^{3} + z^{3}}{xyz} = (1/2)(26)$

$\frac{x^{3} + y^{3} + z^{3}}{xyz} = 13$

Given : $(x-y)^2+(y-z)^2+(z-x)^2=xyz$

On multiplying by two on both sides, we get - $\frac{1}{2}{(x-y)^2+(y-z)^2+(z-x)^2}=\frac{1}{2}xyz$ $\Rightarrow x^2 +y^2 +z^2 -xy -yz -zx = \frac{1}{2}xyz$

Now we multiply by $x+y+z$ on both sides, we get - $(x+y+z)(x^2 +y^2 +z^2 -xy -yz -zx) = \frac{1}{2}(xyz)(x+y+z)$ $\Rightarrow x^3 + y^3 + z^3 - 3xyz = \frac{1}{2}(xyz)(x+y+z)$

But $x+y+z = 20$ , so $\Rightarrow x^3 + y^3 + z^3 - 3xyz = \frac{1}{2}(xyz).20 = 10xyz$ $\Rightarrow x^3 + y^3 + z^3 = 13xyz$ $\frac{x^3 + y^3 + z^3}{xyz} = 13$

$(x+y+z)^2=400$

$x^2+y^2+z^2=400-(2xy+2xz+2yz)$ .....(1)

$(x-y)^2+(y-z)^2+(z-x)^2=xyz$

$2x^2+2y^2+2z^2-2xy-2xz-2yz=xyz$ .....(2)

substitute (1) to (2)

$800-6(xy+xz+yz)=xyz$

$xy+xz+yz=\frac{800-xyz}{6}$

$x^3+y^3+z^3=(x+y+z)^3-3(x+y+z)(xy+xz+yz)+3xyz$

$x^3+y^3+z^3=8000-60(\frac{800-xyz}{6})+3xyz$

$x^3+y^3+z^3=13xyz$

so the answer is $\boxed{13}$ .

From $(x-y)^2+(y-z)^2+(z-x)^2=xyz$ , we have $2(x^2+y^2+z^2-xy-yz-zx)=xyz$

We have: $\frac{x^3+y^3+z^3}{xyz}-3=\frac{x^3+y^3+z^3-3xyz}{xyz}$

$=\frac{(x+y+z)(x^2+y^2+z^2-xy-yz-zx)}{xyz}=20.\frac{1}{2}=10$

Therefore, $\frac{x^3+y^3+z^3}{xyz}=13$

Expand the second equation you will get $x^2+y^2+z^2-xy-yz-xz= xyz/2$

Then multiplies both sides by 20, you will get $(x+y+z)(x^2+y^2+z^2-xy-yz-xz)= 10xyz$

By solving this equation, you will get $x^3+y^3+z^3-3xyz= 10xyz$

Therefore, $(x^3+y^3+z^3)/xyz= 13$

By using the second given condition We get x^2 + y^2 + z^2 - xy - yz -xz = xyz/2 Now using identity x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - yz -xz) We get x^3 + y^3 + z^3 = 13xyz Hence answer is 13.

we know that x^3 + y^3 + z^3 = ( x^2 + y^2 + z^2 - xy - yz - zx ) * ( x+y+z) ...... (1) (x-y)^2 + (y-z)^2 + (z-x)^2 = 2 ( x^2 + y^2 + z^2 - xy - yz - zx ) = xyz ..........(2) PUT x+y+z=20 in eqn (1) ........... x^3 + y^3 + z^3 = 20 * ( xyz / 2 )

Use => a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc). and replace ∑(x-y)^2 with => 2(x^2+y^2+z^2 - xy-yz-zx) .. :) Found it easier than usual level 4 questions..

Multiply both equations together: (x+y+z) [x^2 - 2xy + y^2 + y^2 - 2zy + z^2 + z^2 - 2xz + x^2] = 20xyz (x+y+z)(x^2+y^2+z^2 - xy - yz - xz) = 10xyz x^3 + y^3 + z^3 − 3xyz = 10xyz x^3 + y^3 + z^3 = 13xyz (x^3 + y^3 + z^3) / xyz = 13

Very easy problem for 180 points...

First, let's show some identities...

$\frac {1}{2} (x-y)^2 + (y-z)^2 +(z-x)^2 = x^2+y^2+z^2-xy-yz-zx$

$\Longrightarrow (x-y)^2 + (y-z)^2 +(z-x)^2 = 2(x^2+y^2+z^2-xy-yz-zx)$

And, $(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=x^3+y^3+z^3-3xyz$

Now, let's multiply the two equations in the question...

$(x+y+z){(x-y)^2+(y-z)^2+(z-x)^2} = 20xyz$

$\Longrightarrow (x+y+z) \times 2(x^2+y^2+z^2-xy-yz-zx) = 20xyz$

$\Longrightarrow (x+y+z)(x^2+y^2+z^2-xy-yz-zx)=10xyz$

$\Longrightarrow x^3+y^3+z^3 -3xyz =10xyz$

$\Longrightarrow x^3+y^3+z^3=13xyz$

$\Longrightarrow \frac {x^3+y^3+z^3}{xyz} =13$

Hence, the required answer is $\fbox {13}$

Since x^3+y^3+z^3 - 3xyz = (x+y+z)1/2( (x-y)^2 + (y-z)^2 + (z-x)^2 ) = 20/2 (xyz)= 10xyz so x^3+y^3+z^3 = 13xyz and finally x^3+y^3+z^3/xyz = 13

1/2{(x−y)^2+(y−z)^2+(z−x)^2}=1/2xyz. (x^3+y^3+z^3)/xyz = [1/2xyz(x+y+z)+3xyz]/xyz =13

Note that $x^3 + y^3 + z^3 - 3xyz = \left(x+y+z\right)\left(x^2+y^2+z^2-xy-zy-zx\right)$

From equation ( $2$ ), we have that $xyz = 2\left(x^2+y^2+z^2-xy-zy-zx\right)$

So $x^3+y^3+z^3 = \frac{xyz\left(x+y+z\right)}{2} -3xyz$

From equation ( $1$ ) we have that $x+y+z = 20$

So $x^3 + y^3 + z^3 = 10xyz+3xyz$

Therefore $\frac{x^3+y^3+z^3}{xyz} = \frac{13xyz}{xyz} = 13$

Thus the required answer is $13$

Multiply both equations to get: $(x+y+z)*((x-y)^2+(y-z)^2+(z-x)^2)$ = $20xyz$ $(x+y+z)*(x^2+y^2+z^2-xy-yz-zx)$ = $10xyz$ $x^3+y^3+z^3-3xyz$ = $10xyz$ $x^3+y^3+z^3$ = $13xyz$ $(x^3+y^3+z^3):(xyz)$ = $13$

Expanding: $2x^2+2y^2+2z^2-2xy-2xz-2yz=xyz$ Recall the useful identity: $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2--xy-xz-yz)$ Now, we can find $x^3+y^3+z^3$ : $x^3+y^3+z^3=20\left(\dfrac{xyz}{2}\right)+3xyz=10xyz+3xyz=13xyz$ The answer is $\boxed{13}$ .

First, note that $x^3 + y^3 +z^3 -3xyz = (x+y+z)(x^2+y^2+z^2-xy-xz-yz)$ and that $(x-y)^2 +(x-z)^2 + (y-z)^2 = 2(x^2+y^2+z^2-xy-xz-yz)$ .

Then it follows that

$\frac{x^3+y^3+z^3}{xyz} = \frac{x^3+y^3+z^3-3xyz +3xyz}{xyz} =\\ \frac{(x+y+z)(x^2+y^2+z^2-xy-xz-yz)}{2(x^2+y^2+z^2-xy-xz-yz)} +3 = \frac{20}{2}+3 = 13.$

Expanding $xyz$ , we get

$xyz = 2x^{2} + 2y^{2} + 2z^{2} -2xy - 2xz - 2yz$ (1)

Then,

$x^{2}yz = 2x^{3} + 2xy^{2} + 2xz^{2} - 2x^{2}y - 2x^{2}z - 2xyz$ (2) $xy^{2}z = 2x^{2}y + 2y^{3} + 2yz^{2} - 2xy^{2} - 2xyz - 2y^{2}z$ (3) $xyz^{2} = 2x^{2}z + 2y^{2}z + 2z^{3} - 2xyz - 2xz^{2} - 2yz^{2}$ (4)