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Lesson 7 of 8

Exploring Algebraic Identities · Lesson 7 of 8

Finding New Identities

“Once you recognise the pattern, inventing identities feels suspiciously powerful.”

Learning Objectives

• Derive and visualise cubic binomial identities. • Use sum and difference of cubes. • Understand the three-variable cubic identity. • Apply identities to higher-level algebraic problems. • Simplify rational algebraic expressions by factorisation.

New identities can be discovered by combining identities we already know with distributivity. This is an important mathematical habit: instead of memorising disconnected formulas, we build new results from familiar structures.

The Cube of a Sum

DerivationLaTeX
Cube of a sumLaTeX

Volume Model

A cube of edge a+b has volume (a+b)³. It can be partitioned into two smaller cubes with volumes a³ and b³, three cuboids of volume a²b, and three cuboids of volume ab².

Cube model for (a plus b) cubed A cube of edge a plus b is split into two depth layers. The layer of depth a contains a cubed, two a squared b cuboids and one ab squared cuboid. The layer of depth b contains one a squared b cuboid, two ab squared cuboids and b cubed. (a + b)³ Cube edge = a + b Split into two depth layers Layer of depth a Layer of depth b a³ a²b a²b ab² a b a b a²b ab² ab² b³ a b a b (a + b)³ = a³ + 3a²b + 3ab² + b³
Cube model for (a+b)^3— Show a cube of edge a+b partitioned into a³, b³, three a²b cuboids and three ab² cuboids.

Cube of a Difference

Replace b by −b in the cube-of-a-sum identity.

Cube of a differenceLaTeX
Recognising (p+2q)³

Problem
Factor p³+6p²q+12pq²+8q³.

  1. 1.p³ is p cubed and 8q³=(2q)³.
  2. 2.The middle terms match 3p²(2q) and 3p(2q)².
  3. 3.Therefore the expression is (p+2q)³.
Recognising (2n−5m)³

Problem
Factor 8n³−60n²m+150nm²−125m³.

  1. 1.8n³=(2n)³ and 125m³=(5m)³.
  2. 2.The middle terms match −3(2n)²(5m) and +3(2n)(5m)².
  3. 3.Therefore the expression is (2n−5m)³.

Difference and Sum of Cubes

Difference of cubesLaTeX
Sum of cubesLaTeX

These are factorisation identities. Their quadratic factors look similar, but the middle sign changes.

A Repeating Factor Pattern

Since x²−y² and x³−y³ both contain x−y as a factor, the chapter invites us to investigate higher powers such as x⁴−y⁴ and x⁵−y⁵. For x⁴−y⁴, first use difference of squares: (x²−y²)(x²+y²), and then factor x²−y² further.

Three-Variable Cubic Identity

Three-variable identityLaTeX

This identity comes from full distributive expansion. The mixed terms cancel in pairs, leaving only the cube terms and −3xyz.

Worked Example: Sum of Three Cubes

Problem
If x+y+z=10, xyz=25 and x²+y²+z²=38, find x³+y³+z³.

  1. 1.Use (x+y+z)²=x²+y²+z²+2(xy+xz+yz).
  2. 2.100=38+2(xy+xz+yz), so xy+xz+yz=31.
  3. 3.Use the three-variable cubic identity: 10(38−31)=x³+y³+z³−75.
  4. 4.70=x³+y³+z³−75.
  5. 5.Therefore x³+y³+z³=145.

Simplifying Rational Expressions

The chapter also uses factorisation to simplify rational algebraic expressions. The essential rule is that numerator and denominator must be factorised first. Only common factors may be cancelled, and a cancelled denominator factor must be non-zero.

Worked Example: Simplify by Factorisation

Problem
Simplify (x²−7x+12)/(5x²+5x−100), assuming the denominator is non-zero.

  1. 1.x²−7x+12=(x−3)(x−4).
  2. 2.5x²+5x−100=5(x²+x−20).
  3. 3.x²+x−20=(x−4)(x+5).
  4. 4.So the fraction is [(x−3)(x−4)]/[5(x−4)(x+5)].
  5. 5.Cancel x−4 only because it is a common factor and is non-zero in the allowed domain.
  6. 6.The simplified form is (x−3)/[5(x+5)].
Do Not Cancel Terms

Cancellation works with common factors, not individual terms separated by + or −. Factor first.

Practice Problems

Practice Problems
  1. Expand (a+b)³ using distributivity.
  2. Factor 27b³−1/(64b³).
  3. Factor 64y³+z³/125.
  4. Investigate whether x−y is a factor of x⁴−y⁴.
  5. Factor an expression of the form p³+27q³+r³−9pqr using the three-variable identity where applicable.
  6. Simplify a rational algebraic expression by factorising numerator and denominator first.
  7. Factor n³−n and explain why the result is divisible by 6 for natural n.

Key Takeaways

Key Takeaways

• Cubic identities can be derived from square identities and distributivity. • (a+b)³ and (a−b)³ have coefficients 1,3,3,1. • Sum and difference of cubes have standard factor forms. • The three-variable cubic identity produces extensive cancellation. • Rational expressions are simplified by factorising and cancelling common non-zero factors.

Coming Next

Next, we revise the complete chapter and practise choosing the right identity or factorisation method.

Previous · Lesson 6

Factorisation Without Using Algebra Tiles

Next · Lesson 8

Chapter Summary and Practice