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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
2 years, 10 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. ==" Samuraiwarm Tsunayoshi · 2 years, 10 months ago

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@Samuraiwarm Tsunayoshi สพฐ รอบสอง ปี 57 อ่ะครับ ถ้าจำโจทย์ไม่ผิด ขอบคุณสำหรับคำตอบนะครับ Sanchayapol Lewgasamsarn · 2 years, 10 months ago

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@Samuraiwarm Tsunayoshi sorry for skipping several steps cuz i'm too lazy to type =3= Samuraiwarm Tsunayoshi · 2 years, 10 months ago

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Me not either. :P Abhishek Bisht · 2 years, 10 months ago

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