# Book Isle binomial expansion: map coefficients, exponents and substitution checks

Language: en
Canonical: [Book Isle binomial expansion: map coefficients, exponents and substitution checks](<https://allluckly.cn/en/guides/nasbook-binomial-cube-term-exponent-map/>)
中文：[中文 HTML](<https://allluckly.cn/guides/nasbook-binomial-cube-term-exponent-map/>)
English: [English HTML](<https://allluckly.cn/en/guides/nasbook-binomial-cube-term-exponent-map/>)
Published: 2026-11-11T00:00:00+08:00
Modified: 2026-11-11T09:40:26+08:00
Author: Bison

Expand an original cube from three identical factors, distinguishing cubing from multiplying by 3 and checking missing terms without treating samples as proof.

English translation of the [Chinese original](<https://allluckly.cn/guides/nasbook-binomial-cube-term-exponent-map/>), translated on 2026-11-11. Original publication dates, attribution, code and version notes are retained.

A cubed bracket cannot generally be expanded by cubing its two endpoints alone. Follow an original binomial example in Book Isle, expand the multiplication and map each coefficient and exponent. Unlike summation bounds or function composition, this task retains every intermediate term. The app does not perform symbolic expansion or turn two substitutions into proof for every input.

## Fix the complete bracket and exponent

Record title, edition, chapter/page and a nearby phrase. Our target is `(x+2)^3`, with real x. Cubing means multiplying three identical factors, not `3*(x+2)` or cubing only the 2. Preserve brackets, plus sign, variable and exponent when transcribing; wrapping must not attach the exponent to the next term. If resolution prevents reading it, mark the exponent unresolved rather than guessing 2 or 3 from a later answer. This is self-made material, not an attributed textbook quotation; real questions require their own variable domain and definitions.

## Multiply two factors to locate intermediate terms

First calculate `(x+2)*(x+2)`: the four products are x squared, twice x, twice x and 4, combining as `x^2+4*x+4`. The middle four-x term comes from both cross-products; equal contributions must not be deduplicated. Keep product origins and the combined result on separate lines. Like terms require matching variable powers, not merely containing x. Combining x squared with four x into five x squared changes the expression. Before using a coefficient table, trace distributive multiplication so a later omission can be explained rather than memorising an answer.

## Multiply the third factor and combine six terms

Multiply the previous polynomial by x plus 2 to obtain x cubed, twice x squared, four times x squared, eight x, four x and 8. Retain these six source terms before grouping powers into `x^3+6*x^2+12*x+8`. Powers descend from 3 to 0; the constant corresponds to the zero-power position, without requiring a separate zero-to-zero expression at x zero. Check that the square terms combine to six x squared and linear terms to twelve x. Do not simply attach 2, 4, 8 in their encounter order to four final terms. Check the final constant after resolving the intermediate contributions.

## A coefficient pattern still includes powers of two

The binomial formula’s degree-three combination coefficients 1, 3, 3, 1 are not by themselves the final coefficients of powers of x. The four terms also contain powers of 2 with exponents 0, 1, 2, 3, producing 1, 6, 12, 8. Map combination coefficient, power of x, power of 2 and final term separately; do not shift one term’s square of 2 into the next. In a source’s a-plus-b notation, consistently substitute a=x and b=2. Reversed term order can be equivalent, but every term’s internal pairing must remain intact.

## Use substitution to find omissions without calling it proof

At x=1, the original is three cubed, 27, and the expansion gives 1+6+12+8=27. The wrong control `x^3+8` gives 9 there and fails immediately. At x=minus 2, the original is zero; the correct expansion gives minus 8+24−24+8=0. Passing two samples can reveal transcription errors but does not alone prove identity for every real x; retain the distributive derivation. Testing only zero is weak because many expansions missing middle terms still happen to produce the same constant 8.

## Retain parity-dependent signs in the minus variant

Create a separately labelled control `(x-2)^3`, replacing the second quantity by minus 2. Its expansion is `x^3-6*x^2+12*x-8`: odd powers of minus 2 are negative and its square positive. Neither negate every term nor change only the constant. At x=1 both the bracket and 1−6+12−8 give minus 1. Keep plus and minus versions separate rather than saving the control over the original. If a textbook uses other letters, map roles term by term instead of mixing signs or terms from visually similar versions.

## Troubleshoot lost superscripts and joined minus signs

Compare copied notes with the page for cubes, squares, brackets and minus signs; plain text may flatten superscripts into ordinary digits. If needed, adopt the explicit character notation used here and declare its exponent meaning rather than claiming a native equation editor repaired it. Retrieve unsearchable scans by chapter and nearby phrase before deleting and reimporting. For disagreement, check whether the exponent became a square or cubing became multiplication by 3, then review six-term grouping. Do not alter only the final term to match an answer. Keep uncertain source text unresolved and personal rewrites separate from quotations.

## Exit and return to remap terms to their factors

Exit the reader and return to the target. Cover the expansion and rebuild both multiplication rounds from three factors, explaining each intermediate coefficient’s origins. Check x=1, x=minus 2 and the minus variant at 1, but separate sample checks from algebraic justification. Retain six items: edition/location, expression/domain, factors, intermediate products, combined result and substitution checks. Completion means another reader can reproduce the derivation, not automatic identity proof or a rule applied to arbitrary functions or physical quantities merely because a cube symbol appears. Read any changed operation definitions first.

## Related guides and sources

[Book Isle: Match summation indices and limits to expanded terms](<https://allluckly.cn/en/guides/nasbook-summation-index-limits-expansion/>); [Read function composition with Book Isle: pass the inner output to the outer function](<https://allluckly.cn/en/guides/nasbook-function-composition-inner-outer-order/>).

Checked on November 11, 2026: [Apple public information](<https://itunes.apple.com/lookup?id=6766198898&country=cn&entity=software&timestamp=20261111>) lists Book Isle 1.1.4, requiring iOS / iPadOS 15.6 or later. See the [official App Store](<https://apps.apple.com/cn/app/id6766198898>) for published capabilities; verify specific controls in the installed version.

Method reference: [OpenStax: Binomial Theorem](<https://openstax.org/books/algebra-and-trigonometry-2e/pages/13-6-binomial-theorem>). Samples, passages and controls are original exercises. Sources establish definitions and checking methods rather than new automatic analysis in the app. Manual records may be kept externally, and unknown fields should remain explicit. This article does not claim step-by-step testing on a physical device, private server or every output environment.

