Kvant Math Problem 72
Let
Verified: no
Verdicts: SKIP + SKIP
Solve time: 5m01s
Source on kvant.digital
Problem
Solve the equation $$\sqrt[3]{1-x}+\sqrt[3]{1+x}=3,$$ where $p$ is an arbitrary real number.
Yu. I. Ionin
Exploration
Let
$$a=\sqrt[3]{1-x},\qquad b=\sqrt[3]{1+x}.$$
Then the equation becomes
$$a+b=3.$$
The quantities $a$ and $b$ satisfy
$$a^3+b^3=(1-x)+(1+x)=2.$$
Hence we seek real numbers $a,b$ such that
$$a+b=3,\qquad a^3+b^3=2.$$
Using
$$a^3+b^3=(a+b)^3-3ab(a+b),$$
we obtain
$$2=27-9ab,$$
so
$$ab=\frac{25}{9}.$$
Now $a$ and $b$ would have to be roots of
$$t^2-3t+\frac{25}{9}=0.$$
Its discriminant equals
$$\Delta=9-\frac{100}{9} =-\frac{19}{9}<0.$$
Thus no real numbers $a,b$ satisfy both conditions.
Before accepting this, check whether cube roots could introduce some hidden issue. Real cube roots are defined for all real numbers and satisfy $(\sqrt[3]{y})^3=y$, so the substitution is reversible. Hence the contradiction shows there is no real $x$.
The step most likely to hide an error is the passage from $a+b=3$ and $a^3+b^3=2$ to $ab=25/9$. Recomputing:
$$2=(a+b)^3-3ab(a+b)=27-9ab,$$
indeed gives $ab=25/9$.
Problem Understanding
We must determine all real numbers $x$ satisfying
$$\sqrt[3]{1-x}+\sqrt[3]{1+x}=3.$$
This is a Type A problem, a classification problem. We must find every real solution and prove that no other real solutions exist.
The core difficulty is that the equation contains two cube roots. The natural idea is to replace them by new variables and exploit the identities involving $a+b$ and $a^3+b^3$.
The expected answer is that there are no real solutions, because the conditions imposed on the cube roots force a quadratic equation whose discriminant is negative.
Proof Architecture
Lemma 1. If
$$a=\sqrt[3]{1-x},\qquad b=\sqrt[3]{1+x},$$
then
$$a+b=3,\qquad a^3+b^3=2.$$
This follows directly from the given equation and from $a^3=1-x$, $b^3=1+x$.
Lemma 2. Any real numbers $a,b$ satisfying
$$a+b=3,\qquad a^3+b^3=2$$
must satisfy
$$ab=\frac{25}{9}.$$
This follows from the identity
$$a^3+b^3=(a+b)^3-3ab(a+b).$$
Lemma 3. No real numbers $a,b$ satisfy
$$a+b=3,\qquad ab=\frac{25}{9}.$$
Indeed, $a$ and $b$ would be roots of
$$t^2-3t+\frac{25}{9}=0,$$
whose discriminant is negative.
The hardest direction is proving nonexistence. The lemma most likely to fail under scrutiny is Lemma 2, because an arithmetic mistake there would change the conclusion.
Solution
Set
$$a=\sqrt[3]{1-x},\qquad b=\sqrt[3]{1+x}.$$
The given equation becomes
$$a+b=3.$$
Since real cube roots satisfy $(\sqrt[3]{y})^3=y$, we also have
$$a^3=1-x,\qquad b^3=1+x.$$
Adding these equalities yields
$$a^3+b^3=2.$$
Using the identity
$$a^3+b^3=(a+b)^3-3ab(a+b),$$
and substituting $a+b=3$, we obtain
$$2=27-9ab.$$
Hence
$$ab=\frac{25}{9}.$$
Now $a$ and $b$ are real numbers with sum $3$ and product $25/9$. Therefore they are roots of the quadratic equation
$$t^2-3t+\frac{25}{9}=0.$$
Its discriminant is
$$\Delta=9-\frac{100}{9} =-\frac{19}{9}<0.$$
A quadratic equation with negative discriminant has no real roots. Thus no real numbers $a$ and $b$ can satisfy
$$a+b=3,\qquad ab=\frac{25}{9}.$$
This contradiction shows that no real number $x$ satisfies the original equation.
Therefore the set of real solutions is empty:
$$\boxed{\varnothing}.$$
Verification of Key Steps
The first delicate step is the computation of $ab$. Starting from
$$a^3+b^3=2,$$
and substituting
$$a^3+b^3=(a+b)^3-3ab(a+b),$$
together with $a+b=3$, gives
$$2=27-9ab.$$
Solving,
$$9ab=25,\qquad ab=\frac{25}{9}.$$
Any arithmetic error here would alter the final conclusion.
The second delicate step is the nonexistence argument. From
$$a+b=3,\qquad ab=\frac{25}{9},$$
the polynomial with roots $a,b$ is
$$t^2-3t+\frac{25}{9}.$$
Its discriminant equals
$$(-3)^2-4\cdot1\cdot\frac{25}{9} =9-\frac{100}{9} =-\frac{19}{9}.$$
Since this is negative, there are no real roots. The contradiction is complete.
A separate check comes from the inequality
$$(a+b)^2\ge 4ab$$
for real numbers. Here
$$9=(a+b)^2\ge 4ab=\frac{100}{9},$$
which would imply
$$81\ge 100,$$
impossible. This independently confirms that no real $a,b$ exist.
Alternative Approaches
After obtaining
$$a+b=3$$
and
$$a^3+b^3=2,$$
one may use the identity
$$a^3+b^3=(a+b)(a^2-ab+b^2).$$
Substituting $a+b=3$ gives
$$a^2-ab+b^2=\frac23.$$
Since
$$a^2+b^2=(a+b)^2-2ab=9-2ab,$$
we get
$$9-3ab=\frac23,$$
hence
$$ab=\frac{25}{9}.$$
Applying
$$(a-b)^2=(a+b)^2-4ab$$
yields
$$(a-b)^2=9-\frac{100}{9} =-\frac{19}{9},$$
which is impossible for real numbers.
The main solution is preferable because it reaches the contradiction directly through Vieta's formulas and a discriminant computation.