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Philippine Mathematical Olympiad, National Orals

Algebra Difficulty 5.0 AIME, harder Find the answer Philippines

Problem:
Let x=−2+3+5x = -\sqrt{2} + \sqrt{3} + \sqrt{5}, y=2−3+5y = \sqrt{2} - \sqrt{3} + \sqrt{5}, and z=2+3−5z = \sqrt{2} + \sqrt{3} - \sqrt{5}. What is the value of the expression below?
x4(x−y)(x−z)+y4(y−z)(y−x)+z4(z−x)(z−y) \frac{x^{4}}{(x-y)(x-z)} + \frac{y^{4}}{(y-z)(y-x)} + \frac{z^{4}}{(z-x)(z-y)}

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:
Writing the expression as a single fraction, we have
x4y−xy4+y4z−yz4−zx4+xz4(x−y)(x−z)(y−z) \frac{x^{4} y - x y^{4} + y^{4} z - y z^{4} - z x^{4} + x z^{4}}{(x-y)(x-z)(y-z)}
Note that if x=yx = y, x=zx = z, or y=zy = z, then the numerator of the expression above will be 00. Thus, (x−y)(x−z)(y−z)(x-y)(x-z)(y-z) divides x4y−xy4+y4z−yz4−zx4+xz4x^{4} y - x y^{4} + y^{4} z - y z^{4} - z x^{4} + x z^{4}. Moreover, the numerator can be factored as follows.
(x−y)(x−z)(y−z)(x2+y2+z2+xy+yz+zx) (x-y)(x-z)(y-z)\left(x^{2} + y^{2} + z^{2} + x y + y z + z x\right)
Hence, we are only evaluating x2+y2+z2+xy+yz+zxx^{2} + y^{2} + z^{2} + x y + y z + z x, which is equal to
12[(x+y)2+(y+z)2+(z+x)2]=12[(25)2+(22)2+(23)2]=2(5+2+3)=20 \begin{aligned} \frac{1}{2}\left[(x+y)^{2} + (y+z)^{2} + (z+x)^{2}\right] & = \frac{1}{2}\left[(2 \sqrt{5})^{2} + (2 \sqrt{2})^{2} + (2 \sqrt{3})^{2}\right] \\ & = 2(5 + 2 + 3) \\ & = 20 \end{aligned}

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.