Is it possible to prove this equality without calculator?

Recall that by Binomial Expansion

(a + b)^3 = a^3 + b^3 + 3a^2 b + 3ab^2

(a + b)^3 = a^3 + b^3 + 3ab(a + b)

A = ³√(847 - 342√6) + ³√(847 + 342√6)

A³ = (847 - 342√6) + (847 + 342√6) + 3(847 - 342√6)^(2/3) * (847 + 342√6)^(1/3) + 3(847 - 342√6)^(1/3) * (847 + 342√6)^(2/3)

A³ = 1694 + 3(847 - 342√6)^(2/3) * (847 + 342√6)^(1/3) + 3(847 - 342√6)^(1/3) * (847 + 342√6)^(2/3)

A³ = 1694 + 3(847 - 342√6)^(1/3) * (847 + 342√6)^(1/3) [ ³√(847 - 342√6) + ³√(847 + 342√6) ]

A³ = 1694 + 3 * [ (847 - 342√6) (847 + 342√6) ]^(1/3) * [ ³√(847 - 342√6) + ³√(847 + 342√6) ]

A³ = 1694 + 3 * [ 847² - 342² * 6 ]^(1/3) * [ ³√(847 - 342√6) + ³√(847 + 342√6) ]

A³ = 1694 + 3 * [ 15625 ]^(1/3) * [ ³√(847 - 342√6) + ³√(847 + 342√6) ]

A³ = 1694 + 3 * [ 25^3 ]^(1/3) * [ ³√(847 - 342√6) + ³√(847 + 342√6) ]

A³ = 1694 + 3 * 25 * [ ³√(847 - 342√6) + ³√(847 + 342√6) ]

A³ = 1694 + 75* [ ³√(847 - 342√6) + ³√(847 + 342√6) ]

A³ = 1694 + 75* [ A ]

A³ - 75A - 1694 = 0

By Trial and Error A = 14 is a solution, factorizing it gives

(A - 14)(A² + 14A + 121) = 0

But the Quadratic Equation has a negative discrminant, so it has no real solution

Thus A = 14 only

Doneeeeee

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Ok tell me how to solve x^3 - 75x - 1694 = 0 other than Trial and Error? Cubic Formula? Rational Root Theorem? Bounded Theorem? Synthetic Division? Long Division? Bisection Method? Newton-Raphson Method? Trial and Error is still the simplest. Well obviously, you get x = ³√(847 - 342√6) + ³√(847 + 342√6) by Cubic Formula. Most of the time you just need to use trial and error for polynomial of degree higher than 2.

EDIT: I've haven't mastered Cardano's method, but after reading your solution in the other question, do you think you've simplified or shorten the problem? Even I don't like to use trial and error but sometimes we don't have an alternative way to approach it. Your Cardano's method creates more problems than it solves. Anyway, nice try. And calculus is not based on using calculator....

EDIT2:

A=³√(847 - 342√6) + ³√(847 + 342√6) = (7 - 2√6) + (7 - 2√6) = 14

Let a + b√6 = (847 - 342√6)^(1/3)

==> (a + b√6)^3 = 847 - 342√6

==> (a^3 + 18ab^2) + √6 (3a^2 b + 6b^3) = 847 - 342√6

==> Compare coefficients of terms of √6 and non-√6

==> a^3 + 18ab^2 = 847, 3a^2 b + 6b^3 = -342

But then again, you can't exactly solve the two equations simultaneously without using Trial and Error (solution of a = 7, b = -2). Do the same method for (847 + 342√6). It's still trial and error.

So it still boils down to whether you want to solve

A³ - 75A - 1694 = 0

or

a^3 + 18ab^2 = 847, 3a^2 b + 6b^3 = -342

WITHOUT trial and error

Don't you think the first equation is easier to solve?

Not to mention, assume you want to solve a^3 + 18ab^2 = 847, 3a^2 b + 6b^3 = -342, first you need to make either a or b the subject (in which you can't), then you substitute into another, which turns out to be a more complicated polynomial than A³ - 75A - 1694 = 0

A = ³√(847 - 342√6) + ³√(847 + 342√6)

The cube roots must look like a ± b√6.

Since you know that the sum is equal to 14, then clearly a = 7.

(7 + b√6)^3

= 1(7)^3 + 3(7)^2 (b√6) + 3 (7)(b√6)^2 + 1(b√6)^3

= 343 + 147 b √6 + 126 b^2 + 6b^3 √6

= (343 + 126 b^2) + (147b +6b^3)√b

So (343 + 126 b^2) + (147b +6b^3)√b = 847 + 342√6

343 + 126 b^2 = 847

49 + 18 b^2 = 121

18 b^2 = 72

b^2 = 4

b = 2

147b +6b^3 = 342 is true when b = 2.

So ³√(847 - 342√6) + ³√(847 + 342√6) = (7 - 2√6) + (7 - 2√6) = 14

i don't think so...