Chapter 2.6 — Exercise 2.5 — Algebraic Identities

The Eight Identities

• (x + y)² = x² + 2xy + y²
• (x − y)² = x² − 2xy + y²
• (x + y)(x − y) = x² − y²
• (x + a)(x + b) = x² + (a + b)x + ab
• (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx
• (x + y)³ = x³ + 3x²y + 3xy² + y³ and (x − y)³ = x³ − 3x²y + 3xy² − y³
• x³ + y³ = (x + y)(x² − xy + y²) and x³ − y³ = (x − y)(x² + xy + y²)
• x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² − xy − yz − zx)

Multiplying Without Multiplying

Question 1 asks for products like (x + 5)(x + 2): matching the x² + (a+b)x + ab identity with a = 5, b = 2 gives x² + 7x + 10 immediately. (x − 5)(x − 5) is (x − 5)², expanding to x² − 10x + 25. (3x + 2)(3x − 2) fits the difference-of-squares pattern directly, giving 9x² − 4. (1 + x)(1 + x) fits (x+y)² with x → 1, y → x, giving 1 + 2x + x². Recognising the pattern first and substituting into it afterward is consistently faster than expanding term by term — especially once three or four terms are involved, where a direct expansion has far more places for a sign error to slip in unnoticed.

Squaring and Cubing Round Numbers

Question 2 evaluates products like 101 × 99 without direct multiplication — rewritten as (100 + 1)(100 − 1) = 100² − 1² = 9999. 999 × 999 becomes (1000 − 1)² = 1000000 − 2000 + 1 = 998001. 30.5 × 29.5 becomes (30 + 0.5)(30 − 0.5) = 900 − 0.25 = 899.75.

Question 7 applies the same idea to cubes: 99³ = (100 − 1)³ = 1000000 − 30000 + 300 − 1 = 970299, and 1001³ = (1000 + 1)³ = 1000000000 + 3000000 + 3000 + 1 = 1003003001. In both directions — squares and cubes — the trick is the same: rewrite an inconvenient number as a round number plus or minus something small, then let the identity carry the rest of the arithmetic. 50½ × 49½ works the same way despite involving fractions: written as (50 + ½)(50 − ½), it becomes 50² − ¼ = 2500 − ¼ = 2499¾, no different in structure from the whole-number cases.

Factorising by Spotting the Pattern

Question 3 reverses the process: given an expanded expression, identify which identity produced it. 16x² + 24xy + 9y² matches x² + 2xy + y² with x → 4x, y → 3y, factorising to (4x + 3y)². 18a² − 50 = 2(9a² − 25) = 2(3a + 5)(3a − 5), pulling out a common factor of 2 before the difference-of-squares pattern becomes visible. 3p² − 24p + 36 = 3(p² − 8p + 12) = 3(p − 6)(p − 2), where the trinomial inside the brackets splits via a + b = −8, ab = 12.

Three-Term Squares and Binomial Cubes

Question 4 expands three-term squares like (x + 2y + 4z)² using the six-term identity, giving x² + 4y² + 16z² + 4xy + 16yz + 8zx, and expands binomial cubes like (2a − 3b)³ using the cube identity, giving 8a³ − 36a²b + 54ab² − 27b³. Question 5 runs the equivalent factorisations in reverse: 25x² + 16y² + 4z² − 40xy + 16yz − 20xz matches the three-term square pattern with −5x, 4y, 2z, factorising to (−5x + 4y + 2z)².

Factorising Sums and Differences of Cubes

Question 8 factorises expressions like 8a³ + b³ + 12a²b + 6ab² by first recognising it as (2a)³ + b³ + 3(2a)(b)(2a + b) — exactly the expanded form of (2a + b)³, so it factorises to (2a + b)³ directly, without needing the general sum-of-cubes identity at all. Questions 9 and 10 use that general identity instead: verified numerically first (x = 3, y = 7 confirms x³ + y³ = (x+y)(x²−xy+y²) on both sides), then applied to factorise 27a³ + 64b³ = (3a + 4b)(9a² − 12ab + 16b²) and 343y³ − 1000 = (7y − 10)(49y² + 70y + 100). Question 11 extends the same idea to three terms: 27x³ + y³ + z³ − 9xyz = (3x + y + z)(9x² + y² + z² − 3xy − yz − 3zx).

The Special Case x + y + z = 0

The identity x³ + y³ + z³ − 3xyz = (x+y+z)(x²+y²+z²−xy−yz−zx) collapses to something remarkable whenever x + y + z = 0: the entire right-hand side becomes zero times anything, so x³ + y³ + z³ = 3xyz follows immediately, with no further work. This single fact evaluates otherwise painful-looking sums of three separate cubes almost instantly. (−10)³ + 7³ + 3³: since −10 + 7 + 3 = 0, the answer is just 3(−10)(7)(3) = −630. Similarly, 28³ + (−15)³ + (−13)³ = 3(28)(−15)(−13) = 16380, since 28 − 15 − 13 = 0.

From Algebra to Rectangles and Cuboids

The closing questions connect factorisation to measurement. Given a rectangle's area as 4a² + 4a − 3, splitting the middle term (4a² + 6a − 2a − 3) factorises to (2a + 3)(2a − 1) — one possible length and breadth. Given a cuboid's volume as 3x³ − 12x, factoring out 3x first gives 3x(x² − 4) = 3x(x + 2)(x − 2), three possible dimensions read straight off the three factors. These problems are a reminder that factorisation isn't only an algebra exercise for its own sake — a factored expression can describe physical dimensions just as naturally as an expanded one describes an area or a volume. It's also worth noting that these answers aren't unique in an absolute sense: swapping which factor gets called "length" and which gets called "breadth" describes the same rectangle, since multiplication doesn't care about the order of its factors.

Looking Back Across the Chapter

Every technique across this chapter — classifying by degree, finding zeroes, applying the Remainder and Factor Theorems, and now recognising identities — solves some version of the same underlying question: what does a polynomial look like when it's broken into simpler pieces? Revisiting the introduction is worth doing once this exercise is done, since degree and term-count — the very first ideas in the chapter — are exactly what determine which identity among these eight is even worth trying first. The identities themselves resurface directly when Class 10's quadratic equations use factorisation as one of several methods for solving ax² + bx + c = 0.