mathematical treasure part 2

i have deleted my answer
and letssee now who can solve this[1]

r u sure this is a mathematical trasure ..

just substitute a-x=t³

we will get

a-t³+a³=t

thus

t³-a³+t-a=0

(t-a)(t²+a²-ta)+(t-a)=0

we get one solution as t=a

and for the other

t²+a²-at+1=0

here D=a²-4(a²+1) <0 thus we wont get any real solution here

so only solution

t=a

or ³√a-x=a

thus x=a-a³

perfectly correct
but a more elegant way follows

we have

a=³√³√a-x-x

Let f(f(a)=₃√a-x

we here observe that it is an increasing function as f"(a)>0

f(f(a))=a
this is possible when

f(a)=a

so we get

₃√a-x=a

so

x=a-a3

hence the only solution is a-a3

yes
that's why i posted it
so that we can think that way also[1]