HP50g simplifing a root

[Albert Chan]

(09-29-2020 09:22 PM)peacecalc Wrote:  \[ - 15\sqrt{3} = - 3a^2b - b^3 \]

This is enough to solve it all
Let b = c√3, we have -15√3 = -3a²(c√3) - (c√3)³

c * (a²+c²) = 5       ⇒ c = 1, a = ±2

(±2 - √3)³ = (±8) - 3(4)(√3) + 3(±2)(3) - (3√3) = ±26 - 15√3

→ ³√(26 - 15√3) = 2 - √3

Quote:Look at the more complicate expression for simplification:
\[ \left( 9416 - 4256\sqrt{5} \right)^{\frac{1}{3}} = ? \]

(a + b√5)³ = a³ + 3a²b (√5) + 3ab²(5) + b³ (5√5)

Collect radical free terms: a³ + 15 ab² = a * (a²+15b²) = 9416 = (2³)(11)(107)

(a²+15b²) > 0 ⇒ a > 0

15b² = 9416/a - a², thus RHS must be divisible by 3, and ends in 0 or 5

a=1: 9416/1-1² = 9415, not divisible by 3
a=2: 9416/2-2² = 4704, not end in 0 or 5
a=4: 9416/4-4² = 2338, not end in 0 or 5
a=8: 9416/8-8² = 1113, not end in 0 or 5

a=11: 9416/11-11² = 735 = 15*49 → |b| = 7

This work:

(11 ± 7√5)³ = 1331 + 3(121)(±7√5) + 3(11)(49*5) + (±343)(5√5) = 9416 ± 4256√5

→ ³√(9416 - 4256√5) = 11 - 7√5

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Comment: both examples assumed the root has form a + b√k, all symbols integer.
Assumption might not hold, they might be rationals (and, more likely, irrationals).

For first example, had we solve the radical free terms, this also produce 26

(1 + (±5/3)(√3))³ = 1 + (3)(±5√3/3) + (3)(25/3) + (125/9)(√3) = 26 ± (170/9)(√3)