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# 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 $$x^{5} + y^{5} + z^{5}$$

Note by Sanchayapol Lewgasamsarn
3 years, 8 months ago

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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. =="

- 3 years, 8 months ago

สพฐ รอบสอง ปี 57 อ่ะครับ ถ้าจำโจทย์ไม่ผิด ขอบคุณสำหรับคำตอบนะครับ

- 3 years, 8 months ago

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

- 3 years, 8 months ago

Me not either. :P

- 3 years, 8 months ago