Cancel and see
Sum and difference of cubes
A long product
Multiply (x − y) by (x² + xy + y²). That is 2 × 3 = 6 products, and they look messy. But many of them are opposites: they add up to zero.
After all the opposite pairs cancel, how many terms are left?
What this lesson covers
The idea
x³ – y³ = (x – y)(x² + xy + y²) and x³ + y³ = (x + y)(x² – xy + y²).
A long product
Multiply (x − y) by (x² + xy + y²). That is 2 × 3 = 6 products, and they look messy. But many of them are opposites: they add up to zero.
After all the opposite pairs cancel, how many terms are left?
- 6
- 4
- 2
Cancel the pairs
Tap a term. Its opposite lights up. Tap the opposite to cancel both. Then see what is left.
Two new identities
x³ − y³ = (x − y)(x² + xy + y²) and x³ + y³ = (x + y)(x² − xy + y²).
(x − y)(x² + xy + y²) = x³ + x²y + xy² − x²y − xy² − y³ = x³ − y³ (x + y)(x² − xy + y²) = x³ − x²y + xy² + x²y − xy² + y³ = x³ + y³
Notes
x³ − y³ = (x − y)(x² + xy + y²) and x³ + y³ = (x + y)(x² − xy + y²). Multiply them out and the middle terms cancel in pairs.
Check yourself
How many products do you get when you multiply (x − y) by (x² + xy + y²), before cancelling?
Answer: 6
2 × 3 = 6 products. Four of them cancel in pairs and leave x³ − y³.
x³ + y³ = ?
Take x = 2 and y = 1. What is x³ − y³?
Answer: 7
8 − 1 = 7. Check: (x − y)(x² + xy + y²) = 1 × (4 + 2 + 1) = 7.
m³ − 8 = ?
- (x + y)(x² − xy + y²) — correct. Yes: the sign in the first bracket is +, and the middle term of the second bracket is −xy.
- (x + y)(x² + xy + y²). This leaves terms that do not cancel: it gives x³ + 2x²y + 2xy² + y³.
- (x − y)(x² − xy + y²). This gives x³ − 2x²y + 2xy² − y³, not x³ + y³.
- (m + 2)(m² − 2m + 4). That is m³ + 8, a sum of cubes. Here we have a difference.
- (m − 2)(m² + 2m + 4) — correct. Yes: 8 = 2³, so x = m and y = 2 in x³ − y³ = (x − y)(x² + xy + y²).
- (m − 2)(m² − 2m + 4). In a difference of cubes the middle term of the second bracket is +xy, here +2m.