A Nice Radical Equation Solved in Three Ways

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[carloshuertas]

Another great explanation, SyberMath! I actually solved for x in this easy math problem in my head. Thanks a lot!

[spafon]

A fourth method: Pull the second term of the left over to the right, then cube both sides. After you do some algebra you end up with the equation -1= (1-x)^(2/3)-2*(1-x)^(1/3). You can then set u=(1-x)^(1/3). You get the quadratic equation u^2-2u+1=0. The only solution is u=1. Hence (1-x)^(1/3)=1, thus 1-x=1 and x=0.

[-basicmaths]

(a+b)^3=8
a^3+b^3+3ab(a+b)=8
2+3ab*2=8
6ab=6
ab=1
(ab)^3=1
1-x^2=1, x^2=0
x=0

[italixgaming]

My solution (of course tremendously faster)
The function t t^(1/3) is concave then the graph of the function is under its tangent at the point (1, 1).
Therefore t^(1/3)<1+1/3.(t-1) if t is not equal to 1.
If we write this for t=1+x and t=1-x and we calculate the sum, we have:
(1+x)^(1/3)+(1-x)^(1/3)<2.
So we know that x=0 is our only candidate and since it works, 0 is the only solution of the equation (1+x)^(1/3)+(1-x)^(1/3)=2.

[johngreen]

I did it using only one variable instead of an a and b. Just let a=(1+x)^(1/3) then a^3=1+x or x=a^3 -1 now subbing we have a+(1-(a^3-1))^(1/3)=2 or a + (2-a^3)^(1/3)=2 then transposing the a on the right side and cubing both sides we 2-a^3= 8-12a+6a^2-a^3 which simplifies to 0=6(1-2a+a^2) which yields 0=6(1-a)^2 or a=1 which gives us x=0 as required.

[moeberry]

Do the guess and check method then rationalize the numerator because it is of the form infinity- infinity. After doing so and taking limits as x tends to plus and minus infinity from x=0 the function is tending to 0. And the function takes its maximum height at y=2

[bartolhrg]

Desmos does allow cube roots (and any other root)
You can find it in funcs button
Or write sqrt for √, *cbrt* for ³√ and nthroot for ⁿ√

[fivestar]

From the 3rd method we may notice that lim where x goes 0 of cbrt(1+x) + cbrt(1-x) is ±∞

[jorgelozanoalcaido]

I love how you thougth this problem with three different methods. I was thinking in another one and I found that you can use the variable change x=z^3-1 or x=z^3+1 and you have another easier equation. Great video :)

[tmacchant]

(a+b)³=a³+b³+3ab(a+b) → 8=2+6ab
→ ab=1
a+b=2, ab=1
→
a, b are solution of t²-2t+1=0 →
a=1, b=1

[giuseppemalaguti]

X=0, elevo al cubo entrambi e membri... Risulta dopo semplificazione (1-x^2)^(1/3)=1....cioe x=0

[hebrawi]

In the second line the b in the ab^2 should be squared

[ikwebyi]

arithmetic-geometric mean inequality may be faster to solve this question

[peterromero]

Desmos does let you write cube roots... just type cbrt

[matthewdodd]

Guess and check is a pretty valid method in this case as x=0 reduces it down to simply 1+1=2

[musicsubicandcebu]

By inspection x can be 0. The cube root of x+1 is always increasing because x+1 is always increasing, and the cube root of x-1 is always decreasing because x-1 is always decreasing. Thus there can be only one intersection point, at (0, 2).

[JohnRandomness]

I noticed that 1 + 1 = 2, but was x=0 the only solution? Yes:

Cube the equation to get (1 + x) + (1 - x) + + cuberoot(1-x)] = 8
2 + 3*cuberoot(1-x²)*2 = 8 ==> cuberoot(1-x²) = 1 or x=0.

[liuzzz]

Lamo:D Thank you for let math more easily to learn. we appreciate that!

[musicsubicandcebu]

Thanks for leaving the bottom half of the final page blank. That's a lot of YT annoyance gone.

[tambuwalmathsclass]

I like everything that has to do with Substitution.
I used u = 1+x and v = 1-x