By Ladislav NebeskyÌ

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X1 − x3 . x1 − x4 . x2 − x3 . x2 − x4 . x3 − x4 ) = φ(x4 , −x1 − x2 . x1 − x3 . x1 − x5 . x2 − x3 . x2 − x5 . x3 − x5 ); (c) (c) (c) φ(x5 , x1 x2 + x3 x4 ) = φ(x5 , x1 x3 + x2 x4 ); φ(x5 , x1 x2 + x3 x4 ) = φ(x4 , x1 x2 + x3 x5 ); φ(x5 , x1 x2 + x3 x4 ) = φ(x4 , x1 x3 + x2 x5 ); (d) (d) φ(x5 , x4 ) = φ(x5 , x3 ); φ(x5 , x4 ) = φ(x4 , x3 ); (d) φ(x5 , x4 ) = φ(x2 , x3 ). ) Of these twelve different conditions (b) . . (d) , some one of which we must employ, (or at least some condition not essentially different from it,) the three marked (b) (c) (d) are easily seen to reduce respectively the three functions (b) (c) (d) to the five-valued form (a); they are therefore admissible, but they give no new information.

If then that function be itself of the second order, it must be capable of being put under the form b = b0 + b1 a1 + b2 a1 2 , b0 , b1 , b2 being functions of the forms b0 = (b0 )0 + (b0 )1 a1 , b1 = (b1 )0 + (b1 )1 a1 , b2 = (b2 )0 + (b2 )1 a1 , in which the radicals a1 and a1 have the forms lately found, and (b0 )0 , . . (b2 )1 are rational functions of a1 , a2 , a3 . And on the same supposition, the three roots x1 , x2 , x3 , of that equation must, in some arrangement or other, be represented by the three expressions, xα = b0 = b0 + b1 a1 + b2 a1 2 , xβ = b1 = b0 + ρ3 b1 a1 + ρ23 b2 a1 2 , xγ = b2 = b0 + ρ23 b1 a1 + ρ3 b2 a1 2 , ρ3 retaining here its recent value: which expressions reciprocally will be true, if the following relations, b0 = 13 (xα + xβ + xγ ), b1 a1 = 13 (xα + ρ23 xβ + ρ3 xγ ), b2 a1 2 = 13 (xα + ρ3 xβ + ρ23 xγ ), can be made to hold good, by any suitable arrangement of the roots xα , xβ , xγ , and by any suitable selection of those rational functions of a1 , a2 , a3 , which have hitherto been left undetermined.

I = (β, γ, α)i. In the first case, the function fi is symmetric with respect to the two quantities xα , xβ , and therefore involves them only by involving their sum and product, which may be thus expressed, xα + xβ = −a1 − xγ , xα xβ = a2 + a1 xγ + x2γ , 33 a1 and a2 being symmetric functions of the three quantities x1 , x2 , x3 , namely, the following, a1 = −(x1 + x2 + x3 ), a 2 = x1 x2 + x 1 x3 + x 2 x3 ; so that if we put, for abridgment, a3 = −x1 x2 x3 , the three quantities x1 , x2 , x3 will be the three roots of the cubic equation x3 + a1 x2 + a2 x + a3 = 0.