An algebra problem by Kshetrapal Dashottar
⎩ ⎪ ⎨ ⎪ ⎧ ( x 2 + x y + y 2 ) ( y 2 + y z + z 2 ) ( z 2 + z x + x 2 ) = x y z ( x 4 + x 2 y 2 + y 4 ) ( y 4 + y 2 z 2 + z 4 ) ( z 4 + z 2 x 2 + x 4 ) = x 3 y 3 z 3
Non-zero real numbers x , y and z satisfy the system of equations above. Find ( x y z ) − 1 .
The answer is 27.
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Since xyz is not equal to 0, We can divide the second relation by the first. Observe that
x^4 + x^2y^2 + y^4 = (x^2 + xy + y^2)(x^2 − xy + y^2),
holds for any x, y. Thus we get
(x^2 − xy + y^2)(y^2 − yz + z^2)(z^2 − zx + x^2) = x^2y^2z^2.
However, for any real numbers x, y, we have
x^2 − xy + y^2 ≥ |xy|.
Since x^2y^2z^2 = |xy| |yz| |zx|, we get
|xy| |yz| |zx| = (x^2 − xy + y^2)(y^2 − yz + z^2)(z^2 − zx + x^2) ≥ |xy| |yz| |zx|.
This is possible only if
x^2 − xy + y^2 = |xy|, y^2 − yz + z^2 = |yz|, z^2 − zx + x^2 = |zx|,
hold simultaneously. However |xy| = ±xy. If x^2−xy+y^2 = −xy, then x^2+y^2 = 0 giving
x = y = 0. Since we are looking for nonzero x, y, z, we conclude that x^2 − xy + y^2 = xy
which is same as x = y. Using the other two relations, we also get y = z and z = x. The
first equation now gives 27x^6 = x^3. This gives x^3 = 1/27(since x = 0), or x = 1/3. We
thus have x = y = z = 1/3.
so (xyz)^-1=27