Algebraic identities and factorisation for CDS

Identities are the fastest tool in CDS algebra. A question that looks like heavy simplification, such as a fraction full of 0.87³ and 0.13³, often collapses to a single number once you spot the identity behind it. The paper also likes "if x + 1/x = 4, find x³ + 1/x³" questions, which take ten seconds with the right identity and two minutes without.

You do not need a long list. You need about a dozen identities, known well enough to recognise them in disguise, and the habit of checking an answer by putting in small numbers.

The core identities

The a + b + c = 0 shortcut

If a + b + c = 0, then a³ + b³ + c³ = 3abc. Why: the three-cube identity has (a + b + c) as a factor, so its left side is zero. This is behind questions like (x − y)³ + (y − z)³ + (z − x)³, where the three terms always add to zero.

The x + 1/x family

Put a = x and b = 1/x, so ab = 1. The identities become:

• x² + 1/x² = (x + 1/x)² − 2
• x² + 1/x² = (x − 1/x)² + 2
• x³ + 1/x³ = (x + 1/x)³ − 3(x + 1/x)
• x³ − 1/x³ = (x − 1/x)³ + 3(x − 1/x)
• x⁴ + 1/x⁴ = (x² + 1/x²)² − 2

Remainder and factor theorems

• Remainder theorem: when a polynomial p(x) is divided by (x − a), the remainder is p(a).
• Factor theorem: (x − a) is a factor of p(x) exactly when p(a) = 0.

Why: p(x) = (x − a) × q(x) + r, where r is a constant. Putting x = a makes the first term zero, leaving r. For a divisor like (2x − 1), put x = 1/2.

Factorising quadratics

To factorise x² + bx + c, find two numbers whose product is c and whose sum is b. For ax² + bx + c, split the middle term using two numbers whose product is ac and whose sum is b.

• 2x² + 7x + 3: ac = 6, and 6 + 1 = 7. So 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).

Worked questions

Question 1: If x + 1/x = 4, find x² + 1/x² and x³ + 1/x³.

• x² + 1/x² = 4² − 2 = 14.
• x³ + 1/x³ = 4³ − 3 × 4 = 64 − 12 = 52.

Question 2: If a + b = 7 and ab = 12, find a² + b² and a³ + b³.

• a² + b² = 49 − 24 = 25.
• a³ + b³ = (a + b)³ − 3ab(a + b) = 343 − 252 = 91.
• Check: a = 3 and b = 4 give 9 + 16 = 25 and 27 + 64 = 91.

Question 3: Find the value of (2 − 5)³ + (5 − 1)³ + (1 − 2)³.

• The three bases are −3, 4 and −1, which add to 0.
• So the sum = 3 × (−3) × 4 × (−1) = 36. Direct check: −27 + 64 − 1 = 36.

Question 4: If a + b + c = 6 and a² + b² + c² = 14, find a³ + b³ + c³ − 3abc.

• 36 = 14 + 2(ab + bc + ca), so ab + bc + ca = 11.
• a² + b² + c² − ab − bc − ca = 14 − 11 = 3.
• The expression = 6 × 3 = 18. Check with a = 1, b = 2, c = 3: 1 + 8 + 27 − 18 = 18.

Question 5: Simplify (0.87³ + 0.13³) ÷ (0.87² − 0.87 × 0.13 + 0.13²).

• This is (a³ + b³) ÷ (a² − ab + b²) with a = 0.87 and b = 0.13.
• It equals a + b = 1.

Question 6: If (x − 2) is a factor of x³ − 3x² + kx − 4, find k.

• By the factor theorem, p(2) = 0: 8 − 12 + 2k − 4 = 0.
• 2k = 8, so k = 4.

Numerical shortcuts from identities

• 101 × 99 = 100² − 1 = 9,999
• 998² = (1,000 − 2)² = 1,000,000 − 4,000 + 4 = 996,004
• 105² − 95² = (105 − 95)(105 + 95) = 10 × 200 = 2,000

Practice set

1. If x − 1/x = 3, find x² + 1/x².
2. If a + b + c = 0 and abc = 6, find a³ + b³ + c³.
3. Factorise x² + 7x + 12.
4. Find 105² − 95².
5. If x + 1/x = 3, find x³ + 1/x³.
6. Find the remainder when x³ + 2x² − 5x + 7 is divided by (x − 1).
7. If a − b = 3 and ab = 10, find a² + b² and a³ − b³.
8. If x = 2 + √3, find x + 1/x and x² + 1/x².

Answers

1. 11. 3² + 2.
2. 18. 3abc = 3 × 6.
3. (x + 3)(x + 4). 3 × 4 = 12 and 3 + 4 = 7.
4. 2,000. 10 × 200.
5. 18. 3³ − 3 × 3 = 27 − 9.
6. 5. p(1) = 1 + 2 − 5 + 7.
7. 29 and 117. a² + b² = 9 + 20. a³ − b³ = 27 + 3 × 10 × 3 = 117. Check: a = 5, b = 2.
8. 4 and 14. 1/x = 2 − √3, so x + 1/x = 4, and x² + 1/x² = 16 − 2.

What to do next

• Write the identity table from memory and test each line with a = 1, b = 2
• Practise ten x + 1/x questions until the chain feels automatic
• Use identities in quadratic equations, where α² + β² comes from the same idea
• Revise simplification with an eye out for hidden identities