Standard algebraic identities

Learning outcome

Derive and apply standard identities to expand expressions, calculate efficiently, and recover sums of squares or cubes from limited information.

Concepts and assumptions

All variables here are real numbers. The polynomial identities used below contain no denominators, so no real values are excluded.

An identity is an equation true for every value in its stated domain. For example, (x + 2)² = x² + 4x + 4 holds for every real x. By contrast, x + 2 = 7 is true only when x = 5. Testing a few values may expose a false identity, but does not prove a general one.

Identities follow from distribution. Expanding (a + b)(a + b) gives a² + ab + ba + b². Since ab = ba, the middle terms combine:

(a + b)² = a² + 2ab + b².

Replacing b by −b gives (a − b)² = a² − 2ab + b². Multiplying opposite-sign brackets makes the middle terms cancel:

(a + b)(a − b) = a² − b².

Similarly, (x + a)(x + b) = x² + (a + b)x + ab. Multiply the squared expansion by one more bracket to obtain:

(a + b)³ = a³ + 3a²b + 3ab² + b³.

Replacing b by −b gives (a − b)³ = a³ − 3a²b + 3ab² − b³. Regrouping the middle cubic terms gives a³ + b³ = (a + b)³ − 3ab(a + b).

Identify which quantities occupy the roles of a and b before substituting. An identity works with numbers or entire algebraic expressions.

Worked examples

Example 1 — A squared difference. Expand (2x − 3)².

Here a = 2x and b = 3. (2x − 3)² = (2x)² − 2(2x)(3) + 3² = 4x² − 12x + 9. Both the coefficient and variable are squared in (2x)².

Example 2 — Convenient multiplication. Calculate 1003 × 997.

These numbers lie 3 above and below 1000: (1000 + 3)(1000 − 3) = 1000² − 3² = 1,000,000 − 9 = 999,991.

Example 3 — Using a sum and product. Given a + b = 11 and ab = 24, find a² + b² and a³ + b³.

From the square identity, a² + b² = (a + b)² − 2ab = 121 − 48 = 73. From the cubic relation: a³ + b³ = 11³ − 3(24)(11) = 1331 − 792 = 539. Neither calculation requires finding a and b separately.

Common mistakes

Omitting the middle term in a square; writing (a − b)² as a² − b²; failing to square coefficients; assuming numerical checks prove an identity.

Practice questions

1. Expand (3x + 4)².
2. Calculate 104² using an identity.
3. Given a − b = 5 and ab = 14, find a² + b².
4. Given x + y = 9 and xy = 18, find x³ + y³.

Worked solutions

1. (3x)² + 2(3x)(4) + 4² = 9x² + 24x + 16.
2. (100 + 4)² = 10,000 + 800 + 16 = 10,816.
3. Since (a − b)² = a² + b² − 2ab, the required sum is 5² + 2(14) = 25 + 28 = 53.
4. x³ + y³ = (x + y)³ − 3xy(x + y) = 9³ − 3(18)(9) = 729 − 486 = 243.