Factorisation by identities

Expansion multiplies brackets into a spread-out polynomial. Factorisation by identities runs that process backwards: you look at a spread-out expression, recognise that it is the result of a known expansion, and rewrite it as a product of factors. This sits in the ICSE Class 9 Algebra chapter right after expansions, and it depends directly on them — every factorisation identity here is simply an expansion identity read from right to left. The syllabus fixes exactly two patterns to master: the difference of two squares, $a^2-b^2$ , and the sum or difference of two cubes, $a^3\pm b^3$ . Both are pattern-recognition skills before they are algebra skills — you must first see that an expression fits the shape before any identity can be applied. Get that recognition solid here, because the very next topic, splitting the middle term, exists precisely to handle the quadratics that do not fit either of these two clean patterns.

The core insight

An expression that is a perfect square minus a perfect square, or a perfect cube plus or minus a perfect cube, factorises instantly by matching it to a fixed identity — no trial and error required.

The core idea

Three identities cover everything this topic asks of you, and all three are just familiar expansions reversed.

a^2 - b^2 = (a-b)(a+b)

where: $a$ and $b$ are any two terms whose squares appear, connected by a minus sign.

a^3 + b^3 = (a+b)(a^2-ab+b^2)

a^3 - b^3 = (a-b)(a^2+ab+b^2)

where: in both cube identities, $a$ and $b$ are the cube roots of the two terms; the sign joining $a^3$ and $b^3$ decides which identity applies. Before reaching for any identity, check two things. First, does a common factor run through every term? Pull it out first — it shrinks the numbers you then work with. Second, does the expression actually fit one of the three shapes — two perfect squares joined by a minus, or two perfect cubes joined by a plus or minus, with no extra middle term? A sum of two squares, $a^2+b^2$ , does not factorise this way over ordinary numbers, and forcing the identity onto it is the single most common error in this topic. The two cube identities are mirror images, and the sign pattern is the part students mix up. For $a^3+b^3$ , the linear factor keeps the plus sign, $(a+b)$ , while the quadratic factor carries a minus in its middle term, $(a^2-ab+b^2)$ . For $a^3-b^3$ , it flips: the linear factor has a minus, $(a-b)$ , and the quadratic factor carries a plus, $(a^2+ab+b^2)$ . In both cases the sign on the linear factor and the sign in the middle of the quadratic factor are opposite to each other — that single rule regenerates the whole pattern if memory ever slips. Rather than memorise blindly, it helps to see why the cancellation happens. Expand $(a-b)(a+b)$ term by term: $a^2+ab-ab-b^2$ . The two middle terms are opposite and cancel, leaving exactly $a^2-b^2$ . The cube identities cancel the same way — expanding $(a+b)(a^2-ab+b^2)$ produces six terms, and four of them cancel in pairs, leaving only $a^3+b^3$ . Reconstructing this cancellation once by hand is a faster memory aid than repeating the formula, because you can always ask "which sign makes the middle terms die?" and rebuild the identity from scratch. A single expression can also hide more than one identity, applied in sequence. $x^4-y^4$ is first a difference of squares: $(x^2-y^2)(x^2+y^2)$ . But the first bracket, $x^2-y^2$ , is itself a further difference of squares, so it splits again into $(x-y)(x+y)$ . The full factorisation is $x^4-y^4 = (x-y)(x+y)(x^2+y^2)$ , and stopping after the first split is an incomplete answer — always re-check every resulting factor for a hidden identity.

Figure

A visual proof that a square minus a smaller square equals a rectangle of width sum and height difference

Derivation AFactorising a difference of squares

The goal is to show, using only the distributive law and nothing beyond it, that any expression of the exact form $a^2-b^2$ can always be rewritten as the product $(a-b)(a+b)$ , and to confirm the reverse direction works too, so the identity can be trusted as a genuine two-way equivalence rather than a rule to apply blindly.

Step 1. Start from the product $(a-b)(a+b)$ and expand it using the distributive law, treating $(a-b)$ as a single term multiplying each part of $(a+b)$ : $(a-b)(a+b) = a(a+b) - b(a+b)$ .

Step 2. Distribute each piece fully: $a(a+b) = a^2+ab$ and $b(a+b) = ab+b^2$ , so the expression becomes $a^2+ab-ab-b^2$ .

Step 3. Collect like terms. The two middle terms, $+ab$ and $-ab$ , are equal in size and opposite in sign, so they cancel completely, leaving $a^2-b^2$ .

Step 4. Since $(a-b)(a+b)$ expands to exactly $a^2-b^2$ , the two expressions are identical for every value of $a$ and $b$ , so the equality can be read in either direction — as an expansion rule or, reversed, as the factorisation identity $a^2-b^2=(a-b)(a+b)$ used throughout this topic.

Worked example

Factorise $9x^2 - 25y^2$ fully, then factorise $8p^3+27q^3$ .

Step 1. Check the pattern for the first expression. $9x^2=(3x)^2$ and $25y^2=(5y)^2$ , both perfect squares joined by a minus sign, so this matches $a^2-b^2$ with $a=3x$ and $b=5y$ .

Step 2. Apply the identity. $a^2-b^2=(a-b)(a+b)$ becomes $9x^2-25y^2=(3x-5y)(3x+5y)$ . Check by re-expanding: $(3x)^2-(5y)^2=9x^2-25y^2$ , which matches, and neither bracket hides a further identity, so this is fully factorised.

Step 3. Check the pattern for the second expression. $8p^3=(2p)^3$ and $27q^3=(3q)^3$ , joined by a plus sign, matching $a^3+b^3$ with $a=2p$ and $b=3q$ .

Step 4. Apply the identity. $a^3+b^3=(a+b)(a^2-ab+b^2)$ gives $(2p+3q)\big((2p)^2-(2p)(3q)+(3q)^2\big) = (2p+3q)(4p^2-6pq+9q^2)$ .

Step 5. Verify by re-expanding: $(2p+3q)(4p^2-6pq+9q^2) = 8p^3-12p^2q+18pq^2+12p^2q-18pq^2+27q^3$ , and the four middle terms cancel in pairs, leaving exactly $8p^3+27q^3$ . The quadratic factor $4p^2-6pq+9q^2$ does not split further using these identities, so it is left as is.

So $9x^2-25y^2=(3x-5y)(3x+5y)$ and $8p^3+27q^3=(2p+3q)(4p^2-6pq+9q^2)$ .

Competitive stretch

Factorise $x^4-16$ completely. Recognise $x^4-16=(x^2)^2-4^2$ , a difference of squares, giving $(x^2-4)(x^2+4)$ . The bracket $x^2-4$ is itself a difference of squares, $(x-2)(x+2)$ , but $x^2+4$ is a sum of squares and cannot be split further. Complete factorisation: $x^4-16=(x-2)(x+2)(x^2+4)$ . This tests exactly the "apply twice, then stop correctly" skill from the core idea.

Common traps

Trying to factorise $a^2+b^2$ with the difference-of-squares identity — a sum of two squares does not factorise this way; do not force it.
Swapping the middle sign in the cube identities, writing $a^2+ab+b^2$ for the sum-of-cubes case instead of $a^2-ab+b^2$ , or the reverse for the difference case.
Extracting the wrong cube or square root when identifying $a$ and $b$ — for example, reading $27q^3$ as $(3q^3)$ instead of correctly as $(3q)^3$ .
Stopping after one application of an identity when the resulting bracket itself hides a further difference of squares, as with $x^4-y^4$ or $x^4-16$ .

Ask the tutor: If the sign pattern in the cube identities keeps slipping, ask your AI tutor to walk through the expansion cancellation once more, or give you three quick timed questions to lock it in.

Why does a sum of two squares like x squared plus 4 not factorise the same way

Because there is no pair of real factors whose product expands to give a plain plus sign between two squared terms with no middle term. Expanding $(x-b)(x+b)$ always produces a minus in the middle position after the cross terms cancel — there is no analogous cancellation that produces a plus. This is exactly why the syllabus states the identity only for the minus case; treating $a^2+b^2$ as factorisable this way is the most common error in this topic, so always check the connecting sign before applying the identity.

How do I quickly tell if an expression is a perfect cube like 8p cubed

Take the cube root of the coefficient and the cube root of the variable part separately, then check they recombine correctly. For $8p^3$ , the cube root of $8$ is $2$ and the cube root of $p^3$ is $p$ , so the term is $(2p)^3$ — confirm by cubing $2p$ back: $(2p)^3=8p^3$ , which matches. The trap is treating the exponent carelessly, for instance mistaking $27q^3$ for $(3q^3)$ instead of $(3q)^3$ ; always cube your candidate answer back to check before committing to the identity.

Do I always need to check for a common factor before using an identity

Yes, and it should be your very first move on any factorisation question, not just this topic. If every term in the expression shares a common factor, pull it out first, since it usually turns large, awkward coefficients into the small perfect squares or cubes the identities expect. It also means that after applying an identity, you should look back at the resulting factors to see whether any of them hides a further common factor or a repeated identity, as with $x^4-y^4$ splitting twice.

Factorisation by identities

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads