Cubing a Bracket Without Multiplying It Out Three Times
Learn the product identity for two brackets, the square of three terms, and the cube of a sum and difference, then use them to find cubes from a sum and a product.
Why is $(x + 3)(x + 5)$ worth memorising as a pattern?
Because the answer is built directly out of the two numbers, with no multiplying required.
The middle coefficient is the sum , and the constant is the product . Once you see that, every expansion of this shape becomes a two-second mental step — and, read backwards, it is exactly how factorising will work.
The pattern extends. Squaring three terms follows a related rule, and cubing a bracket follows another. This page covers the second part of the ICSE Class 8 Mathematics chapter on special products, and every identity in it is the answer to a multiplication written down once.
The middle coefficient is the sum , and the constant is the product . Once you see that, every expansion of this shape becomes a two-second mental step — and, read backwards, it is exactly how factorising will work.
The pattern extends. Squaring three terms follows a related rule, and cubing a bracket follows another. This page covers the second part of the ICSE Class 8 Mathematics chapter on special products, and every identity in it is the answer to a multiplication written down once.
Formula
What are the product, trinomial and cube identities?
Four results extend the work of the first part:
The trinomial square in words: the square of each term, plus twice each of the three possible pairs. There are exactly three pairs — , and — so the expansion has six terms in total. Counting the terms before you start is the quickest way to catch an omission.
The cube in words: the cube of each term, plus times each term squared multiplied by the other. Note the coefficients and that in the signs alternate .
Two rearrangements that do most of the work:
These come straight from grouping the middle terms: . They let you find a sum of cubes from a sum and a product alone.
The error these identities exist to prevent. is not . Test , :
The missing is .
The trinomial square in words: the square of each term, plus twice each of the three possible pairs. There are exactly three pairs — , and — so the expansion has six terms in total. Counting the terms before you start is the quickest way to catch an omission.
The cube in words: the cube of each term, plus times each term squared multiplied by the other. Note the coefficients and that in the signs alternate .
Two rearrangements that do most of the work:
These come straight from grouping the middle terms: . They let you find a sum of cubes from a sum and a product alone.
The error these identities exist to prevent. is not . Test , :
The missing is .
How do you use the product identity for mental multiplication?
Write both numbers as a round value plus a small offset, then read off the sum and product of the offsets.
Worked example 1. Multiply by .
Take , , :
Worked example 2 — both offsets negative. Multiply by .
Here and , so and :
The product of the offsets is positive because both are negative, while their sum is negative. Getting one of those two signs right and the other wrong is the usual mistake, and the check is simple: the answer must be a little below .
Worked example 3 — offsets of opposite sign. Multiply by .
Now , , so and :
Worked example 4 — the algebraic form.
Sum , product . A negative constant means the two numbers have opposite signs — a fact worth carrying into the factorising chapter.
Worked example 5 — squaring three terms. Expand .
Taking , , :
- squares:
- the pair doubled:
- the pair doubled:
- the pair doubled:
Six terms, as predicted. Note : the square of the negative term is positive, and only the pair products involving it turn negative.
Worked example 6 — the trinomial square rearranged. If and , find .
Worked example 1. Multiply by .
Take , , :
Worked example 2 — both offsets negative. Multiply by .
Here and , so and :
The product of the offsets is positive because both are negative, while their sum is negative. Getting one of those two signs right and the other wrong is the usual mistake, and the check is simple: the answer must be a little below .
Worked example 3 — offsets of opposite sign. Multiply by .
Now , , so and :
Worked example 4 — the algebraic form.
Sum , product . A negative constant means the two numbers have opposite signs — a fact worth carrying into the factorising chapter.
Worked example 5 — squaring three terms. Expand .
Taking , , :
- squares:
- the pair doubled:
- the pair doubled:
- the pair doubled:
Six terms, as predicted. Note : the square of the negative term is positive, and only the pair products involving it turn negative.
Worked example 6 — the trinomial square rearranged. If and , find .
How do you expand and use cubes?
**Apply the four coefficients , watching the signs.
Worked example 1.** Expand .
Taking , :
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Notice : the coefficient is cubed too. Writing here would be wrong by a factor of four.
**A check at .** The bracket gives , and the answer gives .
Worked example 2 — a difference. Expand .
The signs alternate. Checking at : the bracket gives , and the answer gives .
Worked example 3 — a sum of cubes from a sum and a product. If and , find .
The check. The numbers are and , since and . Then . Correct — and note the identity never required finding them.
Worked example 4 — a difference of cubes. If and , find .
With , : . Correct.
Why the sign flips. In the difference form the correction term is added, not subtracted. That is not arbitrary — expanding gives , and the middle two terms are , so moving them across the equation changes their sign.
Worked example 5 — the reciprocal form. If , find .
Using , , so that :
The reason this form is so common. Because , the product term collapses to just , and a cube that looks forbidding becomes two small numbers subtracted.
Worked example 6 — cubing a number mentally. Find .
The digits appear in the answer itself, which is a pleasing consequence of the offset being exactly .
Worked example 1.** Expand .
Taking , :
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Notice : the coefficient is cubed too. Writing here would be wrong by a factor of four.
**A check at .** The bracket gives , and the answer gives .
Worked example 2 — a difference. Expand .
The signs alternate. Checking at : the bracket gives , and the answer gives .
Worked example 3 — a sum of cubes from a sum and a product. If and , find .
The check. The numbers are and , since and . Then . Correct — and note the identity never required finding them.
Worked example 4 — a difference of cubes. If and , find .
With , : . Correct.
Why the sign flips. In the difference form the correction term is added, not subtracted. That is not arbitrary — expanding gives , and the middle two terms are , so moving them across the equation changes their sign.
Worked example 5 — the reciprocal form. If , find .
Using , , so that :
The reason this form is so common. Because , the product term collapses to just , and a cube that looks forbidding becomes two small numbers subtracted.
Worked example 6 — cubing a number mentally. Find .
The digits appear in the answer itself, which is a pleasing consequence of the offset being exactly .
Exam tip
Exam tip: count the terms before you expand
— the middle coefficient is the sum, the constant is the product. A negative constant means the two numbers have opposite signs.
For mental products, offsets that are both negative give a negative sum and a positive product: .
has exactly six terms — three squares and twice each of the three pairs , , . Count them.
In a trinomial square the square of a negative term is positive: in you get , and only the and pair terms turn negative.
Cubes use the coefficients **, with signs alternating** in .
Cube the coefficient too: , not .
**.** At , the values are and .
Learn the two rearrangements: and — the correction is subtracted in the first and added in the second.
And check every expansion at : gives both ways.
For mental products, offsets that are both negative give a negative sum and a positive product: .
has exactly six terms — three squares and twice each of the three pairs , , . Count them.
In a trinomial square the square of a negative term is positive: in you get , and only the and pair terms turn negative.
Cubes use the coefficients **, with signs alternating** in .
Cube the coefficient too: , not .
**.** At , the values are and .
Learn the two rearrangements: and — the correction is subtracted in the first and added in the second.
And check every expansion at : gives both ways.
Did you know
Where the numbers 1, 3, 3, 1 come from
The coefficients of are , and those of are . Those are not separate facts to memorise — they are two rows of a single triangle of numbers.
Write on the top row. On each row below, begin and end with , and make every inner entry the sum of the two numbers above it:
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The rows give the coefficients of , , , , in turn. So without any multiplication at all you can write
The powers of fall from to while the powers of rise, and the next row of the triangle would give the fifth power the same way.
Why the addition rule works. Each term of comes from multiplying by , which means every coefficient in the new row is fed by two coefficients in the old one — the term that gained an and the term that gained a . Adding the pair above is exactly that bookkeeping.
There is also a neat check on each row: the entries sum to a power of two. Row sums to , and row sums to . Setting in the identity explains why — the left side becomes .
Write on the top row. On each row below, begin and end with , and make every inner entry the sum of the two numbers above it:
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The rows give the coefficients of , , , , in turn. So without any multiplication at all you can write
The powers of fall from to while the powers of rise, and the next row of the triangle would give the fifth power the same way.
Why the addition rule works. Each term of comes from multiplying by , which means every coefficient in the new row is fed by two coefficients in the old one — the term that gained an and the term that gained a . Adding the pair above is exactly that bookkeeping.
There is also a neat check on each row: the entries sum to a power of two. Row sums to , and row sums to . Setting in the identity explains why — the left side becomes .
Key takeaways
Products, trinomial squares and cubes: quick revision
- : middle coefficient is the sum, constant is the product. So and .
- Mental products: ; ; .
- Two negative offsets give a negative sum and a positive product.
- — six terms, three squares and three doubled pairs.
- . The square of the negative term is .
- Rearranged: with gives .
- Cubes use : , with signs alternating in .
- — at both sides give . And .
- Cube the coefficient: . And **** — at they are and .
- : with , this gives , matching .
- : with , this gives , matching . The correction is added here.
- Reciprocal form: since , gives .
- Mental cube: .
- The coefficients come from a triangle where each entry is the sum of the two above, and each row sums to a power of two.
Write out the triangle to six rows from memory, then use its fourth row to expand two cubes without looking anything up — that is the point at which the identities become yours.
- Mental products: ; ; .
- Two negative offsets give a negative sum and a positive product.
- — six terms, three squares and three doubled pairs.
- . The square of the negative term is .
- Rearranged: with gives .
- Cubes use : , with signs alternating in .
- — at both sides give . And .
- Cube the coefficient: . And **** — at they are and .
- : with , this gives , matching .
- : with , this gives , matching . The correction is added here.
- Reciprocal form: since , gives .
- Mental cube: .
- The coefficients come from a triangle where each entry is the sum of the two above, and each row sums to a power of two.
Write out the triangle to six rows from memory, then use its fourth row to expand two cubes without looking anything up — that is the point at which the identities become yours.