Find $x^3+y^3+z^3$, given that $x+y+z=4$, $xy+yz+zx=2$ and $xyz=1$

Question

The real numbers $x,y,z$ satisfy $$x+y+z=4$$ $$xy+yz+zx=2$$ $$xyz=1$$

Then $x^{3}+y^{3}+z^{3}$=??.

It's for sharing a new ideas, thanks:)

Answer 1

Hint: what is $(x+y+z)^3$? Open up that expression, everything you already have will appear, then just make a substituition and go for the kill.

(Assuming you know what $x+y+z$ is, sure)

Answer 2

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

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

$$\implies x^3+y^3+z^3-3xyz=(x+y+z)(\sum x^2-\sum yz)$$

$$\sum x^2-\sum yz=(x+y+z)^2-3(xy+yz+zx)$$

Answer 3

Given that constraints, $x,y,z$ are the roots of the polynomial: $$ p(w) = w^3-4w^2+2w-1 $$ hence: $$\sum_{a\in\{x,y,z\}} a^3 = 4\sum_{a\in\{x,y,z\}} a^2-2\sum_{a\in\{x,y,z\}}a+3 = 4(4^2-2\cdot 2)-2\cdot 4+3=\color{red}{43}$$ by Viète's theorem. Another chance is given by: $$ x^3+y^3+z^3 = \text{Tr}\;\begin{pmatrix}0 & 0 & 1 \\ 1 & 0 & -2 \\ 0 & 1 & 4\end{pmatrix}^3. $$ That matrix $M$ is the companion matrix of $p$, hence its eigenvalues are $x,y,z$. So the eigenvalues of $M^3$ are $x^3+y^3+z^3$ and the trace of a matrix is at the same time the sum of the eigenvalues and the sum of the diagonal elements.

Answer 4

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

$$x^3+y^3+z^3=(x+y+z)((x+y+z)^2-3(xy+yz+zx))+3xyz$$ substituting the corresponding values we get values we get $$x^3+y^3+z^3=(4)((4)^2-3(2))+3(1)$$ $$=40+3=43$$

Answer 5

$\bf{My\; Solution::}$ Given $x+y+z = 4$ and $xy+yz+zx = 2$ and $xyz = 1$

Now Let $t=x,y,z$ be the roots of the equation $$(t-x)\cdot (t-y)\cdot (t-z) =0$$

Then $$t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz =0$$

So we get $$t^3-4t^2+2t-1=0\Rightarrow t^3=4t^2-2t+1$$

So $$\displaystyle \sum_{\bf{cyclic}{\; x,y,z}}t^3 = 4\sum_{\bf{cyclic}{\; x,y,z}}t^2-2\sum_{\bf{cyclic}{\; x,y,z}}t+\sum_{\bf{cyclic}{\; x,y,z}}1$$

$$\displaystyle = 4(x^2+y^2+z^2)-2(x+y+z)+3=4(x+y+z)^2-8(xy+yz+zx)-2\left(x+y+z\right)+3$$

$$\displaystyle x^3+y^3+z^3 = 4\cdot 4^2-8(2)-2\cdot (4)+3 = 43$$