ch1.2.9-cubes-and-cube-root-extraction

Chapter 1.2.9 – Of Cubes, and of the Extraction of Cube Roots

Summary: Euler derives the binomial cube identity and builds a systematic algorithm for extracting cube roots of compound quantities and numbers, parallel to his square-root algorithm.

Sources: chapter-1.2.9

Last updated: 2026-04-28

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The Binomial Cube

Multiplying $(a+b{)}^2=a^2+2ab+b^2$ again by $(a+b)$ gives (§333): (source: chapter-1.2.9)

$$(a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3.$$

The cube contains the cube of each part plus the mixed term $3a^{2}b+3a{b}^2=3ab(a+b)$, i.e. three times the product of the two parts multiplied by their sum.

Numerical Applications

The identity yields a shortcut for cubing numbers split into two parts (§334): (source: chapter-1.2.9)

$$5^3=(3+2{)}^3=27+8+18\cdot 5=125.$$ $$10^3=(7+3{)}^3=343+27+63\cdot 10=1000.$$ $$36^3=(30+6{)}^3=27000+216+540\cdot 36=46656.$$

Cube Root Extraction Algorithm

Given $a^3+3a^{2}b+3a{b}^2+b^3$, find the root (§335–337): (source: chapter-1.2.9)

1. First term: the cube root of the leading term is $a$.
2. Remainder: subtracting $a^3$ leaves $3a^{2}b+3a{b}^2+b^3=(3a^2+3ab+b^2)b$.
3. Trial divisor: $3a^2$ (three times the square of $a$, already known).
4. Quotient $b$: dividing the remainder by the trial divisor gives the next term.
5. Complete divisor: $3a^2+3ab+b^2$ (add $3ab$ and $b^2$ once $b$ is known).
6. Confirm zero remainder; repeat iteratively for more-term roots.

Algebraic Examples

Worked in §338–339: (source: chapter-1.2.9)

$$\sqrt[3]{a^3+12a^2+48a+64}=a+4.$$ $$\sqrt[3]{a^6-6a^5+15a^4-20a^3+15a^2-6a+1}=a^2-2a+1.$$

Numerical Cube Root Extraction

The same algorithm is the foundation of the standard arithmetic procedure (§339): (source: chapter-1.2.9)

$$\sqrt[3]{2197}=13(=10+3).$$ $$\sqrt[3]{34965783}=327(=300+20+7).$$

Related pages

• ch1.2.6-squares-of-compound-quantities
• ch1.2.7-extraction-of-roots-compound
• cube-roots
• algebraic-identities
• roots
• powers-and-exponents