HP50g simplifing a root

[Albert Chan]

(10-11-2020 06:28 PM)Albert Chan Wrote:  We have ³√(36 + 20i√7) = (-3 + i√7) / (-1)^(2/3)

Mathematica gives many "simplifed" forms, here are 2 of them.

(-3+i√7) * (-1-i√3)/2 = (√21+3)/2 + i/2*(3√3-√7) = (√21+3)/2 + i/2*√(34-6√21)

I was curious, given √(34-6√21), how to "simplify" to (3√3-√7) ?

(10-09-2020 05:21 PM)Albert Chan Wrote:  \(\sqrt[3]{A ± \sqrt{R}} = a ± \sqrt{r} \quad ⇒ \quad a = \large\frac{\sqrt[3]{A+\sqrt{R}}
\;+\; \sqrt[3]{A-\sqrt{R}}}{2} \)

We can use the same trick (if LHS is real, for both ±)

\( \sqrt{A ± B\sqrt{c d}} = a\sqrt{c} \,±\, b\sqrt{d} \quad ⇒ \quad
a\sqrt{c} = \large\frac{\sqrt{A+B\sqrt{c d}}\;+\;\sqrt{A-B\sqrt{c d}}}{2} \)

lua> A, B, k = 34, -6, 21
lua> C = B * sqrt(k)
lua> mean = (sqrt(A+C) + sqrt(A-C))/2
lua> mean^2
27

27 = 3*3*3 = a*a*c. We try a=c=3, d=7, so just solve for b:

(3√3 + b√7)² = (27+7b²) + 6b√21 = 34 - 6√21

We have b = -1, thus √(34 - 6√21) = 3√3 - √7

It is nice that XCas do this automatically

XCas> simplify(√(34-6*√(21)))      → 3√3 - √7