# How to find $x^4+y^4+z^4$ from equation?

Please help me. There are equations: $x+y+z=3, x^2+y^2+z^2=5$ and $x^3+y^3+z^3=7$. The question: what is the result of $x^4+y^4+z^4$?

Ive tried to merge the equation and result in desperado. :( Please explain with simple math as I'm only a junior high school student. Thx a lot

## 2 Answers

solution 2: since $$x^4+y^4+z^4=(x^2+y^2+z^2)^2-2x^2y^2-2y^2z^2-2x^2z^2$$ and $$(xy+yz+xz)^2=x^2y^2+x^2z^2+y^2z^2+2xyz(x+y+z)$$ since $$xy+yz+xz=2,xyz=-\dfrac{2}{3}$$ so $$x^2y^2+y^2z^2+x^2z^2=8$$ so $$x^4+y^4+z^4=5^2-2\cdot 8=9$$

• Wow... Such a long solution... I will try... – Pebrianto Nov 8 '13 at 11:13
• thanks a lot... Avtually it's not that hard... :D – Pebrianto Nov 8 '13 at 17:11

use this

since $$x^2+y^2+z^2=(x+y+z)^2-2xy-2yz-2xz\Longrightarrow xy+yz+xz=2$$ $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)=(x+y+z)^3-3(xy+yz+xz)(x+y+z)$$ so $$7-3xyz=27-18\Longrightarrow xyz=-\dfrac{2}{3}$$ use $$x^4+y^4+z^4=(x+y+z)(x^3+y^3+z^3)-(xy+yz+xz)(x^2+y^2+z^2)+xyz(x+y+z)$$