A Radical Equation From X

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This problem is from Nonroutine Problems in Algebra, Geometry and Trigonometry:

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PLAYLISTS 🎵 :

[coreyyanofsky]

at 5:42 you can either proceed as in the video or you can recall the original equation and replace the LHS with the RHS

[ronbannon]

NIce equation. Easy to modify to create more interesting solutions. e.g. has four simple solutions.

[benjaminvatovez]

Thank you for your video. If we let a, b and c be the first, second and third cubic root, then we have:
a+b=c
a³+b³=c³
then so that c=0 or c²=(a+b)²=a²-ab+b² thus ab=0.
Finally, (a=0 thus x=1) or (b=0 thus x=2) or (c=0 thus x=1.5).

[pwmiles]

cbrt(x - 1) + cbrt(x - 2) = cbrt(2x - 3)

Symmetry be my friend! Substitute
x = u + 3/2
cbrt(u + 1/2) + cbrt(u - 1/2) = cbrt(2u)

Cube both sides
2u + 3 cbrt(u + 1/2) cbrt(u - 1/2) (cbrt(u + 1/2) + cbrt(u - 1/2)) = 2u
3 cbrt(u^2 - 1/4) cbrt(u) = 0

Cube again
27(u^2 - 1/4) u = 0

u=0, x=3/2 or
u^2 - 1/4 = 0
u = +/- 1/2
x = 1 or 2

[devondevon]

x=2 seems to be only that work not 3/2 or 1
[cube root (x-1) + cube root ( x-2)]^3= [cube root ( 2x -3) ]^3
let a= cube root (x-1) and be = cube root ( x-2)
x -1 + x -2 + 3 ( x^2 +2 -3x) [ cube root (2x-3)]= 2x-3 ( a^3 + b^3 + 3ab (a+b)) identity
2x-3 + 3x^2 + 6-9x ( cube root (2x-3) = 2x-3

3x^2 + 6 - 9x (cube root (2x-3) = 2x-3 - (2x-3)
3x^2 + 6- 9x (cube root (2x-3) =0 equation P
3x^2 + 6- 9x = 0 divide both sides by cube root (2x-3)
x^2 - 3x + 2 =0
(x-1)(x-2) =0
x =1
and x =2 but only 2 works

Go back to equation P
3x^2 +6-9x (cube root (2x-3) =0
cube root (2x-3) =0 divided both sides by 3x^2 + 6-9x
2x-3=0
2x=3
x = 3/2 but this doesn't work either
So x = 2 Answer as when x =1 and x= 3/2 the equation is not satisfied.

[raghvendrasingh]

cube root (x-1)+cube root (x-2)+cube root (3-2x) = 0

hence x-1+x-2+3-2x = 3 cube root (x-1)(x-2)(3-2x)
because if a+b+c= 0 then a^3+b^3+c^3 = 3abc

(x-1)(x-2)(3-2x) = 0
hence roots are 1, 2 and 3/2

[dorkmania]

Substitution might make things simpler here. The value inside the radical on the RHS is the sum of the respective inside values on the LHS viz. double their average.

[davidseed]

let a=x-1 and b=x-2
cbrt(a) + cbrt(b)= cbrt(a+b)
cube both sides
a+b+3cbrt(ab(a+b)) = a+b
hence
ab(a+b)=0
so one of those factors a, b, a+b must be zero
solutions
a=0. x=1
b=0, x=2
a+b=0, x=3/2
all solutions work in the original equation (noting cbrt(-1)=-1

[scottleung]

Got 'em all! (BTW nice graph at the end.)

[giuseppemalaguti]

Elevo al cubo 2 volte, rimane l'equazione 2x^3-9x^2+13x-6=0..(x-1)(2x^2-7x+6)=0..(x-1)2(x-2)(x-3/2)=0...x=1, x=2, x=3/2

[othman]

There is a well known formula that solves the problem in one line:
a^3+b^3+c^3=3abc where a+b+c=0 this is a general case of the formula:

Which can be proved by summing up the equations P(x)=0 where x=a, b, c
And
Back to the problem:
Denote every radical in the problem a, b and -c the equation becomes a+b+c=0, we also have a^3+b^3+c^3=0 so abc=0 and the rest is very straightforward.

[davidtaran]

y=x-1.5
t=0.5
(y+t)^1/3+(y-t)^1/3=(2y)^1/3
y+t + y-t
3*((2y)^1/3)*(y^2-t^2)^1/3=0
1.(2y)^1/3 = 0
y=0, x1=1.5
2. (y^2-t^2)^1/3=0
y^2-t^2=0
y1=t, x2=2
y2=-t, x3=1