Now consider the expression 8n3 − 60n2m + 150nm2 − 125m3. If you write it in the form (a − b)3, what will be a and b?
Answer: 8n3 − 60n2m + 150nm2 − 125m3 = (2n − 5m)3, so a = 2n, b = 5m.
Step-by-step solution
Given: 8n3 − 60n2m + 150nm2 − 125m3
To find: a and b such that the expression equals (a − b)3
To find: a and b such that the expression equals (a − b)3
Idea: Compare with (a − b)3 = a3 − 3a2b + 3ab2 − b3. The signs +, −, +, − match. Find a and b from the first and last terms, then check the middle ones.
- 8n3 = (2n)3 so a = 2n; 125m3 = (5m)3 so b = 5m.1 mark
- Check: 3a2b = 3(2n)2(5m) = 3 × 4n2 × 5m = 60n2m ✓; 3ab2 = 3(2n)(5m)2 = 3 × 2n × 25m2 = 150nm2 ✓.1 mark
- So the expression = (2n)3 − 3(2n)2(5m) + 3(2n)(5m)2 − (5m)3 = (2n − 5m)3.½ mark
- Hence a = 2n and b = 5m.½ mark
8n³ − 60n²m + 150nm² − 125m³ = (2n − 5m)³, with a = 2n and b = 5m.
Check: n = 3, m = 1: 216 − 540 + 450 − 125 = 1, and (6 − 5)3 = 1 ✓.
Answer to write in the exam
8n3 − 60n2m + 150nm2 − 125m3 = (2n)3 − 3(2n)2(5m) + 3(2n)(5m)2 − (5m)3
= (2n − 5m)3 [a3 − 3a2b + 3ab2 − b3 = (a − b)3]
∴ a = 2n, b = 5m
Common mistakes that cost marks
- Taking b = −5m. In the form (a − b)3, the minus sign is already in the formula, so b = 5m.
- Taking a = 8n or 4n: the cube root of 8n3 is 2n.
- Mixing up the order: a goes with the positive cube term (8n3), b with the negative one (−125m3).
How this can come in the exam
MCQ (1 mark)
x3 − 6x2 + 12x − 8 equals
- (x − 8)3
- (x − 2)3
- (x + 2)3
- (x − 4)3
Show answer
(B) (x − 2)3
a = x, b = 2: 3x2(2) = 6x2, 3x(4) = 12x.
Short answer (2 marks)
Factorise 27y3 − 54y2 + 36y − 8.
Show answer
27y3 = (3y)3, 8 = 23; 3(3y)2(2) = 54y2, 3(3y)(2)2 = 36y (1 mark). So it is (3y − 2)3 (1 mark).Try one yourself
Write 64a3 − 48a2b + 12ab2 − b3 as a cube.
Show answer
(4a)3 − 3(4a)2b + 3(4a)b2 − b3 = (4a − b)3.
More questions like this
- Now let us play with known identities to discover more identities. Try to multiply the following using the distributive property.
- We already know that x2 − y2 = (x − y)(x + y).
Further, we have verified that x3 − y3 = (x − y)(x2 + xy + y2).
Observe that x − y is a common factor of x2 − y2 and x3 − y3.
Do you think x − y is also a factor of x4 − y4?
Note that x4 − y4 = (x2)2 − (y2)2 = (x2 − y2) (x2 + y2).
Can you see how x − y is a factor of x4 − y4?
How about x5 − y5? Does this also have x − y as a factor? - Exploring further, let us multiply (x + y + z) and (x2 + y2 + z2 − xy − xz − yz).
- The sum of three numbers is 10 and their product is 25. The sum of their squares is 38. Try to use the previous identity to find the sum of the cubes of these three numbers.
- Simplify the rational expression x2 − 7x + 125x2 + 5x − 100, assuming that 5x2 + 5x − 100 ≠ 0.