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Algebraic identities · 3 marks

Now let us play with known identities to discover more identities. Try to multiply the following using the distributive property.

  1. 1. (x − y)(x2 + xy + y2)
  2. 2. (x + y)(x2 − xy + y2)
Answer: 1. (x − y)(x2 + xy + y2) = x3 − y3. 2. (x + y)(x2 − xy + y2) = x3 + y3. Both are identities.

Step-by-step solution

Idea: Multiply each term of the first bracket by every term of the second, then cancel like terms with opposite signs. Almost everything cancels, leaving just two cubes.

1. (x − y)(x2 + xy + y2)

  1. x × (x2 + xy + y2) = x3 + x2y + xy2½ mark
  2. −y × (x2 + xy + y2) = −x2y − xy2 − y3½ mark
  3. Add: +x2y and −x2y cancel; +xy2 and −xy2 cancel. Result: x3 − y3.½ mark
  4. Verify with numbers (as suggested): x = 3, y = 2: (1)(9 + 6 + 4) = 19 and 27 − 8 = 19 ✓. x = 5, y = −1: (6)(25 − 5 + 1) = 126 and 125 − (−1) = 126 ✓.
x3 − y3

2. (x + y)(x2 − xy + y2)

  1. Prediction: replacing y by −y in part 1 gives (x + y)(x2 − xy + y2) = x3 − (−y)3 = x3 + y3.½ mark
  2. Check by multiplying: x × (…) = x3 − x2y + xy2; y × (…) = x2y − xy2 + y3.½ mark
  3. The middle terms cancel, leaving x3 + y3.½ mark
x3 + y3
(x − y)(x² + xy + y²) = x³ − y³ and (x + y)(x² − xy + y²) = x³ + y³.

Check: Part 2 with x = 2, y = 1: (3)(4 − 2 + 1) = 9 and 8 + 1 = 9 ✓.

Answer to write in the exam

1.

(x − y)(x2 + xy + y2) = x(x2 + xy + y2) − y(x2 + xy + y2)

= x3 + x2y + xy2 − x2y − xy2 − y3

∴ (x − y)(x2 + xy + y2) = x3 − y3

2.

(x + y)(x2 − xy + y2) = x(x2 − xy + y2) + y(x2 − xy + y2)

= x3 − x2y + xy2 + x2y − xy2 + y3

∴ (x + y)(x2 − xy + y2) = x3 + y3

Common mistakes that cost marks

  • Mixing up the signs: in x3 − y3 the bracket is (x2 + xy + y2); in x3 + y3 it is (x2 − xy + y2).
  • Thinking (x − y)(x2 + xy + y2) = (x − y)3. The cube (x − y)3 has four terms; this product has two.
  • Forgetting to multiply by the second term of the first bracket, which leaves six terms that do not cancel.

How this can come in the exam

MCQ (1 mark)

(a + 2)(a2 − 2a + 4) equals

  1. a3 + 8
  2. a3 − 8
  3. (a + 2)3
  4. a3 + 6a + 8
Show answer

(A) a3 + 8
(x + y)(x2 − xy + y2) = x3 + y3 with y = 2.

Short answer (2 marks)

Factorise 27p3 − 1.

Show answer27p3 − 1 = (3p)3 − 13 (½ mark) = (3p − 1)((3p)2 + 3p + 1) (1 mark) = (3p − 1)(9p2 + 3p + 1) (½ mark).

Try one yourself

Find the product (2m + 3)(4m2 − 6m + 9).

Show answer

This is (x + y)(x2 − xy + y2) with x = 2m, y = 3: 8m3 + 27.

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