Lesson 3 of 12 · Algebra Foundations
Polynomial Identities
Expanding a polynomial by brute force is slow and error-prone. A handful of identities — the square and cube of a sum, the multi-term square, the binomial theorem, and one striking cube identity — let you go straight to the answer, or to just the one term or coefficient you actually need.
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Learning objectives
- Expand (a ± b)² and (a ± b)³ without redoing the multiplication from scratch each time.
- Apply the multi-term square identity (a + b + c)² to move between symmetric sums.
- Use the binomial theorem to extract a single term or coefficient without expanding the whole expression.
- Recognize and apply the a³ + b³ + c³ − 3abc identity, including its striking special case when a + b + c = 0.
The core idea
Polynomial identities are the algebraic analogue of the arithmetic shortcuts from Lesson 1: instead of multiplying everything out and hoping to spot a simplification afterward, you recognize the shape first and substitute directly into a known expansion. The payoff is largest when a problem only asks for part of the expansion — a single coefficient, or a symmetric combination of variables — because the identity gets you there without ever producing the full expanded polynomial.
Binomial squares and cubes
(a ± b)² = a² ± 2ab + b²
(a ± b)³ = a³ ± 3a²b + 3ab² ± b³
Memorize these as shapes, not formulas to re-derive: the coefficients 1, 2, 1 and 1, 3, 3, 1 come directly from the binomial theorem below.
Multi-term square
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
The three-variable generalization of (a+b)². It's the fastest route between "sum of squares" and "sum of pairwise products" when you know the total sum.
Binomial theorem
(a + b)ⁿ = Σₖ₌₀ⁿ C(n, k) · aⁿ⁻ᵏ · bᵏ
The term containing bᵏ has coefficient C(n, k) = n! / (k!(n−k)!). To find one specific term, solve for the k that matches the power you want — no need to write out the rest of the expansion.
The a³ + b³ + c³ − 3abc identity
a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)
When a + b + c = 0, the right side vanishes entirely, giving the striking special case a³ + b³ + c³ = 3abc — a fast route to a cubic sum whenever a problem hands you three quantities that sum to zero.
Worked example 1 — Direct expansion by identity
Problem
Expand (2x − 3)².
Key insight
Match the shape to (a − b)² with a = 2x, b = 3, and substitute directly.
Solution
(2x)² − 2(2x)(3) + 3² = 4x² − 12x + 9.
Takeaway
No multiplication of binomials needed — just substitute into the known pattern.
Worked example 2 — Multi-term square, solved backwards
Problem
If a + b + c = 10 and a² + b² + c² = 40, find ab + bc + ca.
Key insight
(a+b+c)² = a²+b²+c² + 2(ab+bc+ca) — solve for the pairwise-product sum directly.
Solution
10² = 40 + 2(ab+bc+ca) → ab+bc+ca = 30.
Takeaway
This is the three-variable version of the a²+b² shortcut from Lesson 1 — same idea, one more term.
Worked example 3 — Extracting a single binomial-theorem term
Problem
Find the coefficient of x³ in the expansion of (x + 2)⁵.
Key insight
You need the term where the power of x is 3, i.e. 5 − k = 3, so k = 2 — no need to expand the other five terms.
Solution
The term is C(5,2)·x³·2² = 10 · x³ · 4 = 40x³, so the coefficient is 40.
Takeaway
Whenever a problem asks for "the coefficient of" a single term, solve for the matching k first — writing the full expansion is wasted work.
Worked example 4 — Strategic substitution meets the cube identity
Problem
Compute 101³ exactly, without direct multiplication.
Key insight
Write 101 = 100 + 1 and expand using the (a+b)³ identity.
Solution
100³ + 3(100²)(1) + 3(100)(1²) + 1³ = 1,000,000 + 30,000 + 300 + 1 = 1,030,301.
Takeaway
The round-number substitution habit from Lesson 1 extends naturally to cubes once you have the (a+b)³ identity memorized.
Strategy notes
Match the shape before expanding anything
Before multiplying anything out, ask which named identity the expression resembles. If a problem gives you a sum and asks for a related symmetric quantity, you almost certainly want an identity rather than individual values.
"Find the coefficient of..." is a k-solving problem, not an expansion problem
Set up n − k (or whichever exponent matches your term) equal to the target power, solve for k, then compute just that one term.
Common mistakes
Dropping the middle term in a binomial square
(a + b)² ≠ a² + b². This is the single most common algebra error at every level — the cross term 2ab is not optional.
Sign errors in (a − b)³
The signs alternate: a³ − 3a²b + 3ab² − b³. Losing track of a sign on the second or third term is common when working quickly.
Applying a³+b³+c³=3abc without checking a+b+c=0
This simplified identity only holds in the special case a + b + c = 0. Applying it whenever you see three cubed terms, without verifying the condition, produces a wrong answer.
Practice
19 questions across four difficulty tiers. Some ask you to expand an expression — type it in any equivalent form (any order of terms) and it's checked for algebraic equivalence, not exact text. Others ask for a coefficient or numeric value. Up to three tries per question before the solution is shown.