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TRACT XXXII.

Theorem for the cube root of a binomial.

If - - ’/(tf + y'fl) be = p + Vq;

then shall l/{a — s /b) be = p—Vq. And vice versa.

Demon. For, since \/{a + Vb) = p + \/0»therefore - a +• Vb — p 1 + Sp'-^/q + 3pq + ?\/?; ‘hence - a = p 3 + 3pq,and - - V b = 3p z V q + q V q.

Consequently, by subtraction, it is

a - s/b — p 3 - 3p*V9 + 3pq - qV9,

And the cubic root of this is 3 /[a — Vb) =p — »/ q . a, e. d.

Again, if - - i/{a — </b) be ~p — q\

then shall - - %/(a + Vb) be = p + q.

This is proved like the former, by adding, instead of sub-tracting.

Corol. Hence \/(a + Vb) +£/(« — V'b) = 2 p the root of acubic equation.

The necessity or occasion for a rule to extract the cuberoot of such binomials as a-\-Vb and a—sjb, occurred asearly as the first discovery of the rules for solving cubicequations by Tartalea, since one of the rules is expressedin this form, %/{a + V^) +Y\ a ~V^)‘ O n this occasionTartalea invented the following rule, as stated at large inthe next following Tract, on the History of Algebra. Therule is this, and will hold good in-all such cases as will ad-mit of a perfect cubic root. “ Take,” says Tartalea,“ either of the two terms of the binomial, and divide itinto two such parts, that one of them may be a completecube, and the other part exactly divisible by 3 ; then thecube root of the said cubic part will be one term of the re-quired root; and the square root of the quotient arisingfrom the division of ^ of the second part by the said cube