Harder expansions

Three terms (a trinomial)? Group two of them into one, so it becomes a binomial again, expand, and then expand the inner powers.

Example: (1 + x + x²)³ = ((1 + x) + x²)³.

• = (1 + x)³ + 3(1 + x)²x² + 3(1 + x)x⁴ + x⁶.
• (1 + x)³ = 1 + 3x + 3x² + x³.
• 3x²(1 + 2x + x²) = 3x² + 6x³ + 3x⁴.
• 3x⁴(1 + x) = 3x⁴ + 3x⁵.
• Collect like terms: 1 + 3x + 6x² + 7x³ + 6x⁴ + 3x⁵ + x⁶.

Check by putting x = 1: the left side is 3³ = 27 and the coefficients add to 1 + 3 + 6 + 7 + 6 + 3 + 1 = 27. These sums are long, so a quick check like this catches slips in the collecting step.

Approximation. When the small part is tiny, later terms hardly matter, so a few terms give a good estimate.

• (1.02)⁶ = (1 + 0.02)⁶ = 1 + 6(0.02) + 15(0.02)² + ...
• = 1 + 0.12 + 0.006 + (much smaller terms)
• ≈ 1.126 (the exact value is 1.12616...).

Surds. (√5 + √3)⁴ – (√5 – √3)⁴: subtracting keeps only the terms with odd powers of √3, doubled.

• = 2[⁴C₁(√5)³(√3) + ⁴C₃(√5)(√3)³]
• = 2[4 × 5√5 × √3 + 4 × √5 × 3√3] = 2[20√15 + 12√15] = 64√15.

A factor proof. For different integers a and b, a – b is a factor of aⁿ – bⁿ. Write a = (a – b) + b and expand:

• aⁿ = (a – b)ⁿ + ⁿC₁(a – b)ⁿ⁻¹b + ... + ⁿCₙ₋₁(a – b)bⁿ⁻¹ + bⁿ.
• Every term except the last has (a – b) in it. Move bⁿ across: aⁿ – bⁿ = (a – b) × (a whole number).
• Example: 7³ – 3³ = 343 – 27 = 316 = 4 × 79, and 7 – 3 = 4.

Powers inside powers. Expressions like (x/3 + 3/x)⁴ need the theorem once, and then each power like (x/3)³ is worked out on its own. Keep each term's sign, power and number separate until the end; that is where most marks are lost.