8x^3 − 27 = (□x − □)(□x^2 + □x + □) Fill in each blank

2, 3, 4, 6, 9. 8x^3 = (2x)^3 and 27 = 3^3. With a = 2x and b = 3: a^2 = 4x^2, ab = 6x, b^2 = 9, so it is (2x − 3)(4x^2 + 6x + 9).

2x^3 + 16 = □(x + □)(x^2 − □x + □) Fill in each blank

2, 2, 2, 4. Take out the common factor 2 first: 2(x^3 + 8); then the sum of cubes, 8 = 2^3: 2(x + 2)(x^2 − 2x + 4).

3x^3 − 375 factorised completely is 3(x − 5)(x^2 + 10x + 25). True or false?

False. 3x^3 − 375 = 3(x^3 − 125) = 3(x − 5)(x^2 + 5x + 25): the middle is ab = 5x, not 2ab.

Factorisation · Ages 13–14

Factorising sums and differences of cubes

Common factor first, then a³ ± b³; work out a², ab, b² whole.

How to do it

Method

Common factor first, then a^3 ± b^3; work out a^2, ab, b^2 whole

8x³ − 27y³ = (2x − 3y)(4x² + 6xy + 9y²)

Another way

Check with a value

x = 1: 2 − 54 = −52, 2 × (−2) × 13 = −52

In full

Read the two identities backwards: a³ + b³ ≡ (a + b)(a² − ab + b²) and a³ − b³ ≡ (a − b)(a² + ab + b²). First, if every term shares a factor, take it out: 2x³ − 54 = 2(x³ − 27). Second, write the two terms as cubes and find a and b: x³ − 27 = x³ − 3³, so a = x and b = 3; 8x³ − 27y³ = (2x)³ − (3y)³, so a = 2x and b = 3y. Third, put them into the identity, working out a², ab and b² whole: (2x − 3y)(4x² + 6xy + 9y²), not 2x². So 2x³ − 54 = 2(x − 3)(x² + 3x + 9). The second bracket factorises no further (it isn't a perfect square, whose middle would be 6x). The other order works the same way: 64 − x³ = (4 − x)(16 + 4x + x²). Last, expand to check, or put in a value: at x = 1, 2 − 54 = −52 and 2 × (−2) × 13 = −52, the same.

Worked examples

Have a go first, then open each one for the answer and the working.

Example 1
8x³ − 27 = (x − )(x² + x + )
Fill in each blank
Show the answer and working
Answer: 2, 3, 4, 6, 9
8x³ = (2x)³ and 27 = 3³. With a = 2x and b = 3: a² = 4x², ab = 6x, b² = 9, so it is (2x − 3)(4x² + 6x + 9).
Example 2
2x³ + 16 = (x + )(x² − x + )
Fill in each blank
Show the answer and working
Answer: 2, 2, 2, 4
Take out the common factor 2 first: 2(x³ + 8); then the sum of cubes, 8 = 2³: 2(x + 2)(x² − 2x + 4).
Example 3
3x³ − 375 factorised completely is 3(x − 5)(x² + 10x + 25).
True or false?
Show the answer and working
Answer: False
3x³ − 375 = 3(x³ − 125) = 3(x − 5)(x² + 5x + 25): the middle is ab = 5x, not 2ab.

Practice questions to try

Try these 5; the answers are at the end. For a whole worksheet, see VIP worksheets below.

1.x³ + 343 = (x + )(x² − x + )
Fill in each blank
Answer:
2.3x³ + 3 = (x + )(x² − x + )
Fill in each blank
Answer:
3.216 − x³ = (6 − x)( + 6x + x²)
Answer:
4.Factorise 27x³ − 125y³. What do you get?
- A. (3x − 5y)(3x² + 15xy + 5y²)
- B. (3x − 5y)(9x² − 15xy + 25y²)
- C. (3x − 5y)(9x² + 15xy + 25y²)
- D. (3x − 5y)³
Choose the right answer
Answer:
5.2x³ − 432 factorised completely is 2(x − 6)(x² + 6x + 36).
True or false?
Answer:

All the answers

1. 7, 7, 49
2. 3, 1, 1, 1
3. 36
4. C. (3x − 5y)(9x² + 15xy + 25y²)
5. True

Free worksheet: Factorising sums and differences of cubesWorking out, multiple choice and word problems, with answers and working, to print or save as a PDF

VIP practice notes and papers

Aimed at tests and exams: questions on this topic, new every time, to print or save as a PDF.

Practice notes: The key point, worked examples step by step, then questions of the same kind to follow, hints fading
Practice paper: Set out in parts like a school paper, marked, with answers and full working

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