Polynomials • Topic 1 of 4

Polynomial Identities

Identities are equalities that hold for every value of the variable, and CAT uses them as shortcuts for brutal-looking arithmetic. The core set is (a ± b)² = a² ± 2ab + b², a² − b² = (a − b)(a + b), (a ± b)³ = a³ ± 3a²b + 3ab² ± b³, and a³ ± b³ = (a ± b)(a² ∓ ab + b²). Two CAT favourites do real work: (a + b + c)² = a² + b² + c² + 2(ab + bc + ca) lets you jump between the sum of squares and the sum of pairwise products, and a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca) collapses to 3abc the moment a + b + c = 0. The smart move is reverse recognition: see 10609 as 103², or x³ + 1/x³ as a cube of (x + 1/x) minus 3(x + 1/x). Treat every "evaluate this number" as a hidden identity.

✅ Solved examples

1. Evaluate 103² without long multiplication.
(100 + 3)² = 100² + 2·100·3 + 3² = 10000 + 600 + 9 = 10609.
2. Simplify (217² − 183²).
a² − b² = (a − b)(a + b) = (217 − 183)(217 + 183) = 34 × 400 = 13600.
3. If x + 1/x = 4, find x³ + 1/x³.
x³ + 1/x³ = (x + 1/x)³ − 3(x + 1/x) = 4³ − 3·4 = 64 − 12 = 52.
4. If a + b + c = 0, find the value of (a³ + b³ + c³)/(abc).
When a + b + c = 0, a³ + b³ + c³ = 3abc, so the ratio = 3abc/abc = 3.

✏️ Practice — try these, take hints as needed

1. Evaluate 96 × 104 using an identity.
Write as (100 − 4)(100 + 4).
a² − b² with a = 100, b = 4.
10000 − 16.
9984
2. If x − 1/x = 3, find x³ − 1/x³.
(x − 1/x)³ = x³ − 1/x³ − 3(x − 1/x).
So x³ − 1/x³ = 3³ + 3·3.
27 + 9.
36
3. If a + b = 7 and ab = 10, find a³ + b³.
a³ + b³ = (a + b)³ − 3ab(a + b).
7³ − 3·10·7.
343 − 210.
133
4. If a + b + c = 6 and a² + b² + c² = 14, find ab + bc + ca.
(a+b+c)² = a²+b²+c² + 2(ab+bc+ca).
36 = 14 + 2(ab+bc+ca).
Solve for the pairwise sum.
11
5. If x + 1/x = 5, find x² + 1/x².
Square both sides.
(x + 1/x)² = x² + 1/x² + 2.
25 − 2.
23

📝 Topic test — 8 questions

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