ch2.0.8-cubic-surd-rationalization

Ch2.0.8 — Of the Method of rendering the Irrational Formula $\sqrt{a+bx+c{x}^2+d{x}^3}$ Rational

Summary: Extends the rationalization techniques of ch2.0.4-surd-rationalization to cubic radicands; introduces two ansatz methods (root $=f+px$ and root $=f+px+q{x}^2$) when the constant term is a square; develops a bootstrap from any known solution; warns that each operation yields only one new $x$, and many formulas admit only a finite number of solutions.

Sources: chapter-2.0.8

Last updated: 2026-05-09

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The Problem (§112–114)

Find rational $x$ for which

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

has rational $y$. Three structural differences from the quadratic case in ch2.0.4-surd-rationalization:

1. No general solution: the cubic radicand admits no parametric formula. Each application of the method gives only one new value of $x$. (source: chapter-2.0.8, §112)
2. Bootstrap is mandatory: a starting solution must already be known, or be discovered by trial. (§113)
3. Some formulas admit only finitely many solutions — unlike quadratic radicands, where one solution typically generates infinitely many. (§113)

The natural starting point is when $a=f^2$ (a perfect square): then $x=0$ trivially works, providing the seed. (§114)

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Method 1: Two-Term Root (§115, §117)

Given $f^2+bx+c{x}^2+d{x}^3$ (first term square), suppose

$\sqrt{f^2+bx+c{x}^2+d{x}^3}=f+px.$

Squaring: $f^2+bx+c{x}^2+d{x}^3=f^2+2fpx+p^2{x}^2$. The first terms cancel; choosing $p=b/(2f)$ kills the second; dividing by $x^2$ gives $c+dx=p^2$, hence

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

Example (§115): $1+2x-x^2+x^3$. Take root $=1+x$: squares give $1+2x-x^2+x^3=1+2x+x^2$, so $x^3=2x^2$ and $x=2$. The formula evaluates to $1+4-4+8=9=3^2$.

Example (§115): $4+6x-5x^2+3x^3$. Take root $=2+nx$, force $4n=6$, so $n=3/2$. Then $3x^3=(29/4)x^2$, giving $x=29/12$ and root $=45/8$.

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Method 2: Three-Term Root (§116, §117)

Suppose $\sqrt{f^2+bx+c{x}^2+d{x}^3}=f+px+q{x}^2$. Squaring:

$f^2+bx+c{x}^2+d{x}^3=f^2+2fpx+(2fq+p^2)x^2+2pq{x}^3+q^2{x}^4.$

Killing the first three terms requires $p=b/(2f)$ and $q=(c-p^2)/(2f)$. Dividing the remainder by $x^3$:

$\boxed{x=\frac{d-2pq}{q^2}.}$

Example (§116): $1-4x+6x^2-5x^3$. Take root $=1-2x+h{x}^2$. Forcing the third term gives $h=1$. Then $-5x^3=-4x^3+x^4$, so $x=-1$.

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Failure Mode: $f^2+d{x}^3$ (§118)

When the second and third terms are both absent, both methods collapse to $x=0$:

• Method 1 forces $p=0$, hence $d{x}^3=0$.
• Method 2 forces $p=0$ and $q=0$, hence $d{x}^3=0$.

The same obstacle reappears with $f^2+e{x}^4$ in ch2.0.9-quartic-surd-rationalization (§131). For these “lacunary” formulas a separate seed solution is required.

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Bootstrap from a Known Solution (§119–120)

If $a+bx+c{x}^2+d{x}^3$ becomes a square at $x=f$ (with value $g^2$), substitute $x=f+y$. The transformed formula is

$g^2+(b+2cf+3d{f}^2)y+(c+3df)y^2+d{y}^3,$

whose first term is the square $g^2$. Apply Method 1 or Method 2 to $y$, then recover $x=f+y$.

Example (§119): $3+x^3$ with seed $x=1$. Substitute $x=1+y$ to get $4+3y+3y^2+y^3$.

• Method 1 with root $=2+py$: forces $p=3/4$, yields $y=-39/16$, hence $x=-23/16$.
• Method 2 with root $=2+py+q{y}^2$: $p=3/4$, $q=39/64$, yields $y=352/1521$, hence $x=1873/1521$.

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Finitely-Many-Solution Phenomena

Only $x\in \{0,-1,2\}$ for $1+x^3$ (§121)

With seed $x=2$: substitute $x=2+y$ to get $9+12y+6y^2+y^3$.

• Method 1: $p=2$, $y=-2$, $x=0$ — already known.
• Method 2: $p=2$, $q=1/3$, $y=-3$, $x=-1$ — already known. Trying $x=-1+z$ gives $3z-3z^2+z^3$, whose first term vanishes, so neither method applies.

Euler concludes (with grounds for further confirmation) that $1+x^3$ becomes a square only at $x=0,-1,2$. Compare with the cube case in Ch2.0.10 §155.

Prolix Iteration in $1+3x^3$ (§122–123)

Seeds $x=0,-1,2$ are immediate. From $x=1$ (also a seed), Method 1 yields $x=-5/16$ (with $1+3x^3=(61/64{)}^2$), and Method 2 yields $x=-629/1323$. Starting instead from $x=2$ gives $x=8/25$ via Method 1 and again $x=-629/1323$ via Method 2. Iteration produces increasingly cumbersome fractions and surprisingly does not recover the small known solution $x=2$ from the $x=1$ seed — Euler frankly calls this “an imperfection of the present method.” (§123)

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Factor First When Possible (§124–126)

Some formulas dissolve trivially once factored.

Example (§124): $1-x-x^2+x^3=(1-x{)}^2(1+x)$. Since the formula must be a square and $(1-x{)}^2$ is already one, the cofactor $1+x$ must be a square. Set $1+x=n^2$: $x=n^2-1$ produces a square for every $n$. So $n=2\Rightarrow x=3\Rightarrow$ formula $=16$; $n=3\Rightarrow x=8\Rightarrow$ formula $=441$; etc.

Rule (§125): Always check first whether $a+bx+c{x}^2+d{x}^3=0$ has rational roots (necessarily divisors of $a$ by rational-root-theorem). Each rational root $f$ contributes a factor $(f-x)$.

Special case (§126): When both $a=0$ and $b=0$, the formula reduces to $c{x}^2+d{x}^3=x^2(c+dx)$. The square factor $x^2$ leaves only $c+dx=n^2$ to satisfy: $x=(n^2-c)/d$ — a complete parametric solution.

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Alternative Form of Method 1 (§127)

If $p$ is not fixed to remove the second term, Method 1 instead leads to the quadratic

$b+cx+d{x}^2=2fp+p^2{x}^2⟹x=\frac{p^2-c+\sqrt{p^4-2c{p}^2+8dfp+c^2-4bd}}{2d}.$

Now the question is to make the quartic $p^4-2c{p}^2+8dfp+c^2-4bd$ a square — which falls under Chapter IX. This shows the cubic and quartic radicand problems are intertwined.

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Related pages

• ch2.0.4-surd-rationalization
• ch2.0.9-quartic-surd-rationalization
• ch2.0.10-cubic-formula-as-cube
• rationalization
• indeterminate-analysis
• rational-root-theorem
• cubic-equations