Expanding (a + b)³
Watch how a cube of a binomial expands into four terms using the standard identity.
Equation
y = (x + 2)^3 - (x^3 + 0*x^2*2 + 3*2^2*x + 2^3)
Graph
Table
| x | y |
|---|---|
| -5 | 150 |
| -4 | 96 |
| -3 | 54 |
| -2 | 24 |
| -1 | 6 |
| 0 | 0 |
| 1 | 6 |
| 2 | 24 |
| 3 | 54 |
| 4 | 96 |
| 5 | 150 |
What this lesson covers
What you do
You shape the function y = (x + a)^3 - (x^3 + k*x^2*a + 3*a^2*x + a^3) and watch the graph answer.
Challenges to clear
- (x+a)³ needs a 3x²a term: slide k until the difference is 0 at x = 2.
- Identity: 0 at x = −1 too — the expansion has exactly four terms.
- Harder cube: (2x + a)³. Fix a = 4 first.
- Now the missing x² coefficient. It is 3·(2x)²·a ÷ x²a = 12, not 3 — squaring the 2 changes everything. Slide k until the difference is 0 at x = 2.
Check yourself
Why does the graph staying flat at y=0 prove the identity?
What is a common mistake when expanding (x + a)^3?
If a=2, what is the constant term in the expansion of (x+2)^3?
- Because the graph is a straight line
- Because LHS - RHS = 0 means LHS = RHS for all x and a — correct
- Because the coefficients are all positive
- Because x is always greater than a
- Forgetting the middle terms with coefficient 3 — correct
- Adding the exponents instead of multiplying
- Squaring instead of cubing
- Dividing by 3
- 2
- 4
- 6
- 8 — correct
Think about it
- (x+a)³ = x³ + ?·x²a + 3xa² + a³. What is the missing coefficient?