Eighteen products
Identity for x³ + y³ + z³ − 3xyz
A giant product
Multiply (x + y + z) by (x² + y² + z² − xy − xz − yz). The first bracket has 3 terms and the second has 6, so there are 3 × 6 = 18 products!
How many of the 18 products will cancel in pairs?
What this lesson covers
The idea
(x + y + z)(x² + y² + z² – xy – xz – yz) = x³ + y³ + z³ – 3xyz.
A giant product
Multiply (x + y + z) by (x² + y² + z² − xy − xz − yz). The first bracket has 3 terms and the second has 6, so there are 3 × 6 = 18 products!
How many of the 18 products will cancel in pairs?
- 6
- 12
- 18
Cancel the pairs
Tap a term and then its opposite to cancel the pair. Keep going until no more pairs are left.
Only six stay
(x + y + z)(x² + y² + z² − xy − xz − yz) = x³ + y³ + z³ − 3xyz.
Twelve of the eighteen products cancel in six pairs. The six that stay are x³, y³, z³ and three copies of −xyz, which make −3xyz.
Notes
(x + y + z)(x² + y² + z² − xy − xz − yz) = x³ + y³ + z³ − 3xyz. If x + y + z = 0, then x³ + y³ + z³ = 3xyz.
Check yourself
For x = 1, y = 2, z = 3, find x³ + y³ + z³ − 3xyz.
Answer: 18
1 + 8 + 27 − 3 × 6 = 36 − 18 = 18. Check with the other side: (1 + 2 + 3)(1 + 4 + 9 − 2 − 3 − 6) = 6 × 3 = 18.
If x + y + z = 0, what is the value of x³ + y³ + z³ − 3xyz?
Answer: 0
One factor is 0, so the whole product is 0.
If x + y + z = 0, which is true?
Three numbers have sum 10, product 30 and the sum of their squares is 38. Then xy + xz + yz = 31. Find the sum of their cubes.
Answer: 160
10 × (38 − 31) = 70, so x³ + y³ + z³ = 70 + 3 × 30 = 160.
- x³ + y³ + z³ = 3xyz — correct. Yes. The identity gives x³ + y³ + z³ − 3xyz = 0.
- x³ + y³ + z³ = 0. Try x = 1, y = 2, z = −3: the cubes add to 1 + 8 − 27 = −18, not 0.
- x³ + y³ + z³ = −3xyz. With x = 1, y = 2, z = −3: the cubes add to −18 and 3xyz = −18, so the sign is +.