Help me- An exam to choose the olympiad representatives for Thailand

Help me, I can't solve this yet.

Let $x + y + z = x^{2} + y^{2} + z^{2} = x^{3} + y^{3} + z^{3} = 5$

Find the value of $x^{5} + y^{5} + z^{5}$

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

From $2(xy+yz+zx) = (x+y+z)^{2} - (x^{2} + y^{2} + z^{2}) = 20$

$xy+yz+zx = 10 <1>$

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

$xyz = 10 <2>$

$(1) = (2); xy+yz+zx = xyz$

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1 <3>$

$(\frac{1}{x} + \frac{1}{y} + \frac{1}{z})^2 = 1$

$\frac{1}{x^{2}} + \frac{1}{y^{2}} + \frac{1}{z^{2}} + \frac{x + y + z}{5} = 1$

$\frac{1}{x^{2}} + \frac{1}{y^{2}} + \frac{1}{z^{2}} =0 <4>$

$(x^{2} + y^{2} + z^{2})(x^{3} + y^{3} + z^{3}) = x^{5} + y^{5} + z^{5} + (xy)^{2}(x+y) + (yz)^{2}(y+z) + (zx)^{2}(z+x)$

$25 = x^{5} + y^{5} + z^{5} + 100(\frac{5-x}{x^{2}} + \frac{5-y}{y^{2}} + \frac{5-z}{z^{2}})$

$x^{5} + y^{5} + z^{5} = 25 - 100(\frac{5-x}{x^{2}} + \frac{5-y}{y^{2}} + \frac{5-z}{z^{2}})$

$= 25 - 100(\frac{5}{x^{2}} + \frac{5}{y^{2}} + \frac{5}{z^{2}} - 1)$

$= 25 - 100(-1)$

$= \boxed{125}$ ~~~!

Where's the test come from? I haven't seen this question before. =="

Samuraiwarm Tsunayoshi - 7 years, 1 month ago

sorry for skipping several steps cuz i'm too lazy to type =3=

Samuraiwarm Tsunayoshi - 7 years, 1 month ago

สพฐ รอบสอง ปี 57 อ่ะครับ ถ้าจำโจทย์ไม่ผิด ขอบคุณสำหรับคำตอบนะครับ

Sanchayapol Lewgasamsarn - 7 years, 1 month ago

Me not either. :P

Abhishek Bisht - 7 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$