ch2.0.10-cubic-formula-as-cube

Ch2.0.10 — Of the Method of rendering rational the irrational Formula $\sqrt[3]{a+bx+c{x}^2+d{x}^3}$

Summary: Treats the cube-root analog of the squaring problem — finding rational $x$ for which $a+bx+c{x}^2+d{x}^3$ is a perfect cube. Three subclasses by which end is a cube; bootstrap from a known seed; the famous Fermat-style impossibility for $1+x^3$ being a cube; an example showing how factorable formulas with a square factor admit a parametric cube transformation; a concluding sketch of the biquadrate problem.

Sources: chapter-2.0.10

Last updated: 2026-05-09

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The Problem (§147–148)

Find rational $x$ for which

$a+bx+c{x}^2+d{x}^3=z^3$

has a rational cube root. Three preliminary observations:

1. The technique is bounded at degree 3 in $x$, in contrast to the squaring problem which extends to degree 4 (see ch2.0.9-quartic-surd-rationalization §146). (§147)
2. The reduced case $a+bx=p^3$ is trivial: $x=(p^3-a)/b$. (§147)
3. Without a cube structure on either end, a seed solution is mandatory. (§148)

Three structural cases exist, parallel to those of ch2.0.9-quartic-surd-rationalization:

Case	Form	Ansatz
1	$f^3+bx+c{x}^2+d{x}^3$ (first term cube)	Root $=f+px$
2	$a+bx+c{x}^2+g^3{x}^3$ (last term cube)	Root $=p+gx$
3	$f^3+bx+c{x}^2+g^3{x}^3$ (both cubes)	Two solutions from above plus a third with root $=f+gx$

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Case 1: First Term is a Cube (§149)

Suppose $\sqrt[3]{f^3+bx+c{x}^2+d{x}^3}=f+px$. Cubing: $f^3+3f^{2}px+3f{p}^2{x}^2+p^3{x}^3$. Killing the linear term forces $p=b/(3f^2)$; dividing the rest by $x^2$:

$\boxed{x=\frac{c-3f{p}^2}{p^3-d}.}$

The trivial seed $x=0$ is always available, yielding via this method the first new value.

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Case 2: Last Term is a Cube (§150)

Suppose root $=p+gx$. Cubing: $p^3+3p^{2}gx+3g{p}^2{x}^2+g^3{x}^3$. Killing the quadratic term forces $p=c/(3g^2)$; the remaining linear comparison gives

$\boxed{x=\frac{a-p^3}{3g{p}^2-b}.}$

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Case 3: Both Ends are Cubes (§151)

Beyond the two values from Cases 1 and 2, take root $=f+gx$. Cubing: $f^3+3f^{2}gx+3f{g}^2{x}^2+g^3{x}^3$; the constant and cubic terms cancel, the rest divided by $x$:

$\boxed{x=\frac{b-3f^{2}g}{3f{g}^2-c}.}$

So Case 3 yields three distinct new values per pass.

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Bootstrap from a Known Seed (§152)

If $a+bh+c{h}^2+d{h}^3=k^3$, substitute $x=h+y$ to get a Case 1 formula

$k^3+(b+2ch+3d{h}^2)y+(c+3dh)y^2+d{y}^3,$

then apply Case 1 to $y$.

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Worked Examples

$1+x+x^2$: bootstrapping from $x=0$ (§153)

Case 1 with root $=1+px$. Cubing forces $3p=1$, so $p=1/3$. The remaining terms give $3p^2+p^{3}x=1$, hence

$x=\frac{1-3p^2}{p^3}=\frac{2/3}{1/27}=18,1+18+324=343=7^3.$

Iterating: substitute $x=18+y$, get $343+37y+y^2$; root $=7+py$ forces $p=37/147$; $21p^2+p^{3}y=1$ gives

$y=-\frac{147(147^2-21\cdot 37^2)}{37^3}=-\frac{1049580}{50653},$

with further iterations possible.

$2+x^2$ at $x=5$ (§154)

Seed: $2+25=27=3^3$. Substitute $x=5+y$ to get $27+10y+y^2$. Root $=3+py$ with $p=10/27$ leads to

$y=-\frac{27(27^2-9\cdot 100)}{1000}=-\frac{4617}{1000},x=\frac{383}{1000},$

so $2+x^2=2146689/1000000=(129/100{)}^3$.

$4+x^2$ at $x=2$ and $x=11$ (§158–159)

Two known seeds. Starting at $x=2$: Case 1 gives $x=11$ (the other seed). Starting at $x=11$ with root $=5+py$, $p=22/75$:

$y=-\frac{122625}{10648},x=-\frac{5497}{10648}.$

The Möbius substitution $x=(2+11y)/(1+y)$ (sending $y=0\to 2$, $y=\infty \to 11$) converts the formula into

$\frac{8+60y+177y^2+125y^3}{(1+y{)}^2},$

whose numerator (after multiplying by $1+y$) is to be made a cube. Both endpoints $(2)$ and $(5y)$ contribute cube structure; using Case 2 with root $=p+5y$ and $p=59/25$:

$y=\frac{80379}{367875}.$

The variant $x=(2+11y)/(1-y)$ with root $=2-5y$ (Case 3) gives the dramatic value $x=-1090/27$, with the formula equal to $1191016/729=(106/9{)}^3$.

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Impossibility Results

$1+x^3$ Only at $x=0,-1$ (§155)

The formula $1+x^3$ admits the obvious cubes $x=0$ ($1=1^3$) and $x=-1$ ($0=0^3$). Although Case 3 applies, the natural ansatz root $=1+x$ recovers only these. Substituting $x=-1+y$ leads to $3y-3y^2+y^3$ (Case 2 form), which forces $y=1/0=\infty$ — no new value.

Euler appeals to the theorem (footnote 81) that

The sum of two cubes can never be a cube — i.e., $t^3+x^3=z^3$ has no nontrivial integer solutions.

This is Fermat’s Last Theorem at exponent $n=3$, which Euler himself proved (essentially) in Vollständige Anleitung zur Algebra and earlier work. See sum-of-two-cubes.

$x^3+2$ Only at $x=-1$ (§156)

Substitute $x=-1+y$: $1+3y-3y^2+y^3$. All three cases yield $y=0$ or $y=\infty$, confirming uniqueness.

Productive Counterexample: $3x^3+3$ (§157)

At $x=-1$: trivial ($0=0^3$). At $x=2$: $24+3=27=3^3$. Substituting $x=2+y$: $27+36y+18y^2+3y^3$. Root $=3+py$ with $p=4/3$ gives

$y=-\frac{54}{17},x=-\frac{20}{17},3+3x^3=-\frac{9261}{4913}={\left(-\frac{21}{17}\right)}^3.$

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Biquadrate Sketch (§160)

To make $a+bx+c{x}^2$ a fourth power: first make it a square (with rationalization rule from ch2.0.4-surd-rationalization), then make the resulting square root also a square.

Worked example — $x^2+7$ as a fourth power: rationalizing gives $x=(q^2-7p^2)/(2pq)$, with the square root $(7p^2+q^2)/(2pq)$. Multiply by $2pq$: need $2pq(7p^2+q^2)$ to be a square. Set $q=pz$, divide by $p^4$: need $2z(7+z^2)$ a square. Seed $z=1$. Substitute $z=1+y$, root $=4+5y/2$:

$y=1/8,z=9/8,q=9,p=8,x=367/144.$

Then $7+x^2=279841/20736=(529/144{)}^2=(23/12{)}^4$.

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Free-Parameter Cubes via a Square Factor (§161)

If a formula has the form $c{x}^2$ alone, set root $=px$: then $c{x}^2=p^3{x}^3$ gives $x=c/p^3$. Equivalently, $x=c{q}^3$ for any rational $q$ (writing $q=1/p$).

More generally, any formula of the shape $a(b+cx{)}^2$ admits a free parametric cube root: set root $=(b+cx)/q$. Cubing and dividing by $(b+cx{)}^2$:

$a=\frac{b+cx}{q^3}⟹\boxed{x=\frac{a{q}^3-b}{c}}\text{(any rational }q\text{)}.$

This “shows how useful it is to resolve the given formulas into their factors, whenever it is possible” — the explicit motivation for the next chapter, on factorization techniques.

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Method-Failure Summary

Formula	Why methods fail
$f^3+d{x}^3$ (no middle terms)	Case 1 forces $p=0$, hence trivial $x$
$a+g^3{x}^3$ (no middle terms)	Case 2 forces $p=0$, hence trivial $x$
Neither end a cube, no seed	Bootstrap impossible — must search for seed by trial

These limits parallel the failure modes for square formulas in ch2.0.8-cubic-surd-rationalization §118 and ch2.0.9-quartic-surd-rationalization §131.

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Related pages

• ch2.0.8-cubic-surd-rationalization
• ch2.0.9-quartic-surd-rationalization
• ch2.0.4-surd-rationalization
• sum-of-two-cubes
• rationalization
• indeterminate-analysis
• cubic-equations
• cube-roots