Two cubes, one bracket
Take 5³ − 3³. That is 125 − 27 = 98. Can 98 come from two brackets multiplied?
Step 1 of 5
- Multiply (x − y) by (x² + xy + y²).
- Row one: x times each part.
- Row two: −y times each part.
- Each diagonal pair adds to 0.
- What is left is x³ − y³.
The rule
Yes: (5 − 3)(25 + 15 + 9) = 2 × 49 = 98.
Rule: x³ − y³ = (x − y)(x² + xy + y²). x³ + y³ = (x + y)(x² − xy + y²).
Here x and y are any two numbers.
The second line comes from the first. Put −y in place of y. Then x − (−y) becomes x + y. The term xy becomes −xy. And x³ − (−y)³ becomes x³ + y³.
The last column shows a mistake: cubing x − y or x + y.
| x, y | Cubes | The two brackets | The slip, one bracket cubed |
|---|---|---|---|
| 5, 3 | 5³ − 3³ = 98 | (5 − 3)(25 + 15 + 9) = 98 | (5 − 3)³ = 8 |
| 4, 3 | 4³ − 3³ = 37 | (4 − 3)(16 + 12 + 9) = 37 | (4 − 3)³ = 1 |
| 4, 3 | 4³ + 3³ = 91 | (4 + 3)(16 − 12 + 9) = 91 | (4 + 3)³ = 343 |
Factoring a cube
To factor is to write as brackets multiplied. Write each part as a cube. Then use the rule.
27a³ − 8b³is(3a)³ − (2b)³. Here x = 3a and y = 2b. So it is(3a − 2b)(9a² + 6ab + 4b²).p³ + 125isp³ + 5³. So it is(p + 5)(p² − 5p + 25).
Pick x and y. Compare the left, the right and the slip.
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