Algebraic identities — (a+b)², (a−b)², (a+b)(a−b), (a+b)³, (a−b)³, (a³+b³), (a³−b³): Worked examples

The parent note built the seven identities. Here we do the opposite of memorising: we hunt for every kind of situation these identities can be dropped into, then work one example for each. By the end you should be unable to meet a case you have not already seen.

Before we start, one plain-language reminder of the toolbox (no symbol used before it is named):

Recall The seven tools in one glance

$(a + b)^{2} = a^{2} + 2 ab + b^{2}$ — square of a sum
$(a - b)^{2} = a^{2} - 2 ab + b^{2}$ — square of a difference
$(a + b) (a - b) = a^{2} - b^{2}$ — difference of squares
$(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$ — cube of a sum
$(a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}$ — cube of a difference
$a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})$ — sum of cubes
$a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})$ — difference of cubes

Here the letters $a$ and $b$ are just placeholders: any number, letter, or whole chunk of an expression can slide into their spot. The whole skill is seeing what plays the role of $a$ and what plays the role of $b$ .

The scenario matrix

Every problem these identities can throw at you falls into one of these cells. We will hit each one below.

#	Case class	What is tricky about it	Example that hits it
C1	Both terms positive, simple	The friendly base case	Ex 1
C2	A term is negative (subtraction)	The middle sign flips	Ex 2
C3	A term has a coefficient ( $3 x$ , $2 y$ )	You must square/cube the number too	Ex 3
C4	A term is zero / degenerate	Identity collapses — check it still works	Ex 4
C5	Reverse direction: factoring	You recognise a pattern, not expand it	Ex 5
C6	Cube identity with negatives	Alternating signs, easy to miscount	Ex 6
C7	Sum of cubes factoring	Trinomial with $- ab$	Ex 7
C8	Difference of cubes factoring	Trinomial with $+ ab$ (opposite sign)	Ex 8
C9	Clever numeric shortcut (mental maths)	Choose $a, b$ so the arithmetic is easy	Ex 9
C10	Real-world word problem	Translate words → an identity	Ex 10
C11	Exam twist: given $a + b$ and $ab$ , find $a^{2} + b^{2}$	Rearranging an identity backwards	Ex 11

The worked examples

Worked example Example 1 — Cell C1: both terms positive

Expand $(x + 5)^{2}$ .

Forecast: Guess the three terms before reading on. How big is the middle term?

Match the pattern: here $a = x$ and $b = 5$ . Why this step? The identity only works once we know who is $a$ and who is $b$ .
Write the template: $a^{2} + 2 ab + b^{2}$ . Why this step? We picked the square-of-a-sum tool because the bracket is a sum raised to power 2; this template saves us from multiplying $(x + 5) (x + 5)$ term by term every time.
Substitute: $x^{2} + 2 (x) (5) + 5^{2} = x^{2} + 10 x + 25$ . Why this step? $2 \times x \times 5 = 10 x$ and $5^{2} = 25$ .

Verify: Multiply directly with FOIL — this stands for First, Outer, Inner, Last, the four products you get when multiplying two brackets: First $x \cdot x = x^{2}$ , Outer $x \cdot 5 = 5 x$ , Inner $5 \cdot x = 5 x$ , Last $5 \cdot 5 = 25$ . Adding: $x^{2} + 5 x + 5 x + 25 = x^{2} + 10 x + 25$ . ✓ (FOIL Method is just the Distributive Property applied twice.)

Worked example Example 2 — Cell C2: a negative term

Expand $(x - 6)^{2}$ .

Forecast: Which term becomes negative — the middle one, the last one, or both?

Match: $a = x$ , $b = 6$ , and the sign between them is minus. Why this step? A minus sign means we use the difference tool, whose middle term is $- 2 ab$ . Note we take $b = 6$ (a positive number) and let the identity's built-in minus sign handle the subtraction — that way we never have to juggle a negative $b$ by hand.
Template: $a^{2} - 2 ab + b^{2}$ . Why this step? We chose the square-of-a-difference template because the bracket is a difference raised to power 2; its middle term already carries the minus, so we just slot our $a$ and $b$ in.
Substitute: $x^{2} - 2 (x) (6) + 6^{2} = x^{2} - 12 x + 36$ . Why this step? Only the middle term is negative. The last term is $b^{2} = 6^{2} = 36$ , which is positive because the template squares $b$ — and any real number squared (whether we had thought of $b$ as $+ 6$ or $- 6$ ) is positive. Using $b = 6$ throughout keeps this unambiguous.

Verify: $(x - 6) (x - 6) = x^{2} - 6 x - 6 x + 36 = x^{2} - 12 x + 36$ . ✓ Sanity check with numbers: put $x = 10$ . Left: $(10 - 6)^{2} = 16$ . Right: $100 - 120 + 36 = 16$ . ✓

Worked example Example 4 — Cell C4: a degenerate / zero input

Use the difference-of-squares identity on $(x + 0) (x - 0)$ , and separately on $(a + a) (a - a)$ .

Forecast: Does the identity break when one term is zero? Guess before checking.

First case: $b = 0$ . Identity gives $x^{2} - 0^{2} = x^{2} - 0 = x^{2}$ . Why this step? Setting $b = 0$ should collapse the sum and difference back to plain $x$ . Indeed $(x) (x) = x^{2}$ . The identity survives the degenerate case.
Second case: take $a$ and $b$ equal, so the difference $a - b = 0$ . Here $(a + a) (a - a) = (2 a) (0) = 0$ . Why this step? A product with a zero factor is $0$ . The identity predicts $a^{2} - a^{2} = 0$ — same answer. Good.
Limiting behaviour: as the two terms get closer, $a^{2} - b^{2} = (a + b) (a - b)$ shrinks toward $0$ because the factor $(a - b)$ shrinks toward $0$ . Why this step? This tells you why difference-of-squares is small when two numbers are nearly equal — the "gap" factor does the shrinking.

Verify: With $a = 7, b = 7$ : $(a + a) (a - a) = 14 \times 0 = 0$ and $a^{2} - b^{2} = 49 - 49 = 0$ . ✓

Worked example Example 5 — Cell C5: factoring (identity run backwards)

Factor $49 x^{2} - 81$ .

Forecast: Is this a difference of squares, of cubes, or neither? Look at the powers.

Test for two perfect squares: $49 x^{2} = (7 x)^{2}$ and $81 = 9^{2}$ , joined by a minus. Why this step? Difference of squares needs exactly this shape: square minus square. (See Difference of Squares in Number Theory, Factoring Polynomials.)
Set $a = 7 x$ , $b = 9$ , and read the identity right to left: $a^{2} - b^{2} = (a + b) (a - b)$ . Why this step? We chose difference-of-squares (not a cube tool) because both pieces are squares and they are subtracted; reading it right-to-left turns expanding into factoring.
Write: $49 x^{2} - 81 = (7 x + 9) (7 x - 9)$ . Why this step? Factoring is just the same identity used in reverse — recognising the pattern instead of building it.

Verify: Expand back: $(7 x + 9) (7 x - 9) = 49 x^{2} - 63 x + 63 x - 81 = 49 x^{2} - 81$ . ✓

Worked example Example 8 — Cell C8: difference of cubes factoring

Factor $8 x^{3} - 125$ .

Forecast: Same cubes idea as Ex 7, but a minus joins them. Which sign changes in the trinomial?

Spot two perfect cubes: $8 x^{3} = (2 x)^{3}$ and $125 = 5^{3}$ , joined by minus. Why this step? Difference-of-cubes needs cube minus cube — exactly this shape.
Set $a = 2 x$ , $b = 5$ . Identity: $a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})$ . Why this step? We chose difference-of-cubes because both pieces are cubes and they are subtracted. Contrast with Ex 7: here the trinomial carries $+ ab$ (a plus), the opposite sign from sum-of-cubes. The binomial factor is $(a - b)$ , matching the minus outside.
Substitute:
- $a - b = 2 x - 5$
- $a^{2} = 4 x^{2}$ , $ab = 10 x$ , $b^{2} = 25$ , so the trinomial is $4 x^{2} + 10 x + 25$ .
Answer: $8 x^{3} - 125 = (2 x - 5) (4 x^{2} + 10 x + 25)$ .

Verify: Expand $(2 x - 5) (4 x^{2} + 10 x + 25) = 8 x^{3} + 20 x^{2} + 50 x - 20 x^{2} - 50 x - 125 = 8 x^{3} - 125$ . ✓

Worked example Example 10 — Cell C10: real-world word problem (geometric)

A square garden of side $x$ metres has a smaller square flower-bed of side $y$ metres removed from a corner region so that the leftover L-shape has area $x^{2} - y^{2}$ . A gardener wants to re-lay this L-shape as a single rectangle. What whole-number dimensions work if $x = 8$ m and $y = 5$ m?

Figure description (alt-text): Two panels joined by an orange arrow. Left panel: a large square of side $x = 8$ drawn in ink outline; its top-right $y \times y = 5 \times 5$ corner is hatched in burnt orange and labelled "y (removed)"; the remaining L-shaped region is shaded translucent teal and labelled "L-shape area = x² − y²". Right panel: that same teal region redrawn as one long rectangle, its horizontal side labelled in plum "x + y = 13", its vertical side "x − y = 3", and its interior "area = (x+y)(x−y) = 39 m²". The picture shows the L-shape being cut and slid straight so its long edge becomes $x + y$ and its short edge becomes $x - y$ .

Forecast: The identity $x^{2} - y^{2} = (x + y) (x - y)$ turns an area into a product. What two lengths pop out?

Leftover area $= x^{2} - y^{2} = (x + y) (x - y)$ . Why this step? A product of two lengths is a rectangle's area — the identity hands us the two sides directly.
The rectangle has sides $x + y$ and $x - y$ : that is $8 + 5 = 13$ m by $8 - 5 = 3$ m. Why this step? Look at the right panel of the figure: the L-shape (teal) is cut and slid so the long side becomes $x + y$ and the short side $x - y$ .
Area check: $13 \times 3 = 39$ m $^{2}$ , and $x^{2} - y^{2} = 64 - 25 = 39$ m $^{2}$ .

Verify: $8^{2} - 5^{2} = 39$ and $(8 + 5) (8 - 5) = 13 \times 3 = 39$ . Units: metre $\times$ metre $=$ m $^{2}$ . ✓

Recall Quick self-test (reveal after answering)

Which identity, and which cell, for $121 - m^{2}$ ? ::: Difference of squares (C5): $(11 + m) (11 - m)$ . In $(a - b)^{3}$ , what is the sign of the $3 a b^{2}$ term? ::: Positive ( $+ 3 a b^{2}$ ) — Cell C6. Factor $27 - t^{3}$ : which trinomial sign? ::: Difference of cubes (C8): $(3 - t) (9 + 3 t + t^{2})$ — the trinomial has $+ 3 t$ . Given $a + b = 5$ and $ab = 6$ , what is $a^{2} + b^{2}$ ? ::: $25 - 12 = 13$ (Cell C11).