Simplifying a Nice Radical Expression with Cube Roots

If you need to post a picture of your solution or idea:
#ChallengingMathProblems #RadicalExpressions

⭐ Similar videos:

PLAYLISTS 🎵 :

[premkumarsr]

I solved on my own and it exactly matched with 2nd method. All credits go to U. May GOD bless U

[hans-rudigerdrzimmermann]

Two solutions:
A) recall the algebraic identity a^3 + b^3 + c^3 = 3abc if a+b+c=0. Now a and b are the two roots on the left side and you see easily that a*b= -1 and a+b=x.
So a+b - x = 0. Now plug in c=-x in above equation leading to: x^3 + 3x -14= 0. Factorizing gives x=2 and two other unvalid complex roots because x must be rational. The solution of the question is x=2
B) Set a + b = 1 + z root (f) and cube both sides, then compare coefficients yielding a= 1+root(2). Since a*b=-1 we get b= 1-root(2). So the answer again is a+b=2

[XJWill]

The first radical denests to 1 + sqrt(2) and the second radical denests to 1 - sqrt(2), so the sum is 2.
I have posted methods for denesting radicals in previous videos, so I did not include the details in this comment.

[NadiehFan]

In your first method you are using the following theorem:

If a, b, c, d are rational numbers and √b and √d are irrational then a + √b = c + √d implies a = c and b = d

This is often taken for granted when denesting nested roots, but it *is* a theorem, therefore it should be proved. Clearly, this does not apply if √b and √d are rational (take a = 2, b = 4, c = 3, d = 1, then a+√b = c+√d but a ≠ c and b ≠ d). The proof of this theorem is straightforward:

a + √b = c + √d

implies

√b = (c − a) + √d

Squaring both sides gives

b = (c − a)² + d + 2(c − a)√d

and subtracting (c−a)²+d from both sides then gives

(b − d) − (c − a)² = 2(c − a)√d

Now the left hand side is rational, therefore the right hand side must also be rational. But √d is irrational, therefore the right hand side can only be rational if c − a = 0 which implies a = c. With c−a = 0 the equation (b−d)−(c−a)² = 2(c−a)√d becomes b − d = 0 which implies b = d and this completes the proof.

[alexserdukov]

The input looks like a result of the cubic formula

[echowang]

This actually looks like Cardano’s Cubic Formula!

[notlin]

I don't know more adjectives to post here.
I made it for the 2nd method, but the 1st one is really more interesting.

🇧🇷🇧🇷🇧🇷🇧🇷Brazil🇧🇷🇧🇷🇧🇷🇧🇷

[Biologymus]

Please explain why these 2 numbers are conjugate. Adding your numbers make it equal to 2a.

[MrLidless]

This is the Cardano solution to x(x²+3)=14.

[damiennortier]

Have you try to simplify a "n" root of a imaginary number? Since sqrt i = {sqrt2} / 2 + I * {sqrt 2} / 2, it's possible

[satrajitghosh]

Writing a for (7 + 5√2) ^(1/3)
and b for (7 -5√2) ^(1/3) one gets
a^3 + b^3 = 14 and ab = -1
Hereby (a+b)^3
= a^3 + b^3 + 3ab *(a+b)
Hereby a+b is a solution of
x^3 = 14 - 3x
or x^3 + 3x -14 = 0
or (x+2)(x^2 - 2x + 7) = 0
Hereby x = -2, 1 + i*√6, 1- i*√6

[idealtic]

Wow this question has just got in the test to a gifted high school in Vietnam this morning

[scottleung]

I also got x=2 as the only real solution!