Solving Equation $x+a^3=\sqrt[3]{a-x}$ for Real a

[kaliprasad]

Solve the equation $x+a^3=\sqrt[3]{a-x}$ where a is real

 

[anemone]

My "lame" solution:

Spoiler

By observation, note that $x=a-a^3$ is a real solution for the equation $x+a^3=\sqrt[3]{a-x}$.

Now, if we rewrite the given equation by raising it the to third power and rearrange the terms in descending powers of $x$ and factor it since $x=a-a^3$ is a real solution, we have

$x^3+3a^3x^2+(3a^6+1)x+a(a^8-1)=(x-a+a^3)(x^2+kx+m)$ where $k,\,m$ are constants.

Equating the constant terms from both sides gives $m=\dfrac{a(a^8-1)}{a(a^2-1)}=a^4+a^4+a^2+1$

Equating the coefficients of powers of $x^2$ gives $k=2a^3+a$.

Hence,
$x^3+3a^3x^2+(3a^6+1)x+a(a^8-1)=(x-a+a^3)(x^2+(2a^3+a)x+a^4+a^4+a^2+1)$

And the quadratic formula tells us the other two complex roots for the original equation are

$x=\dfrac{-(2a^3+a)\pm\sqrt{-3a^2-4}}{2}$.

 

[Albert1]

Solve the equation $x+a^3=\sqrt[3]{a-x}---(1)$ where a is real

Spoiler

let :$a-x=b^3,\,\, or \,\, x=a-b^3$
we have :$a^3-b^3+a-b=0$
or $(a-b)(a^2+ab+b^2+1)=0$
since :$a^2+ab+b^2+1=(a+\dfrac {b}{2})^2+\dfrac {3b^2}{4}+1>0$
$\therefore a=b ,\,\, or \,\, x=a-a^3$
is the real solution of (1)
takig the result,from anemone :
$x^3+3a^3x^2+(3a^6+1)x+a^9-a=0-----(2)$
suppose the solutions of (2) are :x=$(a-a^3),y,z$
using Vieta's formula we have:
$a-a^3+y+z=-3a^3$---(3)
$(a-a^3)yz=-(a^9-a)---(4)$
and $y,z$ can be found (with respect to a)from (3) and(4)
both of $y\,\, and \,\, z$ wll be complex as anemone mentioned earlier

 

[juantheron]

Spoiler

Solve the equation $x+a^3=\sqrt[3]{a-x}$ where a is real

Consider function $f(a) = x + a^3$, then $f^{-1}(a) = \sqrt[3]{a - x}$. Hence \[f^{-1}(a) = f(a)\] This can happen if and only if \[a = f(a) = f^{-1}(a)\] i.e. \[a = \sqrt[3]{a - x} = x + a^3\] So \[\boxed{x = a - a^3}\]