Mathematics worksheet for Grade 8 students on factoring sum or difference of two cubes, provided by Heisenberg Tutorial Center.

Let's solve each problem step by step using the sum or difference of two cubes factoring formulas:

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🔷 Factoring Formulas:

- Sum of Two Cubes:
$$
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
$$

- Difference of Two Cubes:
$$
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
$$

We will apply these formulas to factor each expression completely.

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1) $ x^3 + 8 $

Recognize:
$ 8 = 2^3 $, so this is $ x^3 + 2^3 $

Use sum of cubes:
$$
x^3 + 8 = (x + 2)(x^2 - 2x + 4)
$$

✔ Answer: $ (x + 2)(x^2 - 2x + 4) $

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2) $ a^3 + 64 $

$ 64 = 4^3 $, so $ a^3 + 4^3 $

Use sum of cubes:
$$
a^3 + 64 = (a + 4)(a^2 - 4a + 16)
$$

✔ Answer: $ (a + 4)(a^2 - 4a + 16) $

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3) $ a^3 + 216 $

$ 216 = 6^3 $, so $ a^3 + 6^3 $

Use sum of cubes:
$$
a^3 + 216 = (a + 6)(a^2 - 6a + 36)
$$

✔ Answer: $ (a + 6)(a^2 - 6a + 36) $

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4) $ 27 + 8x^3 $

Rewrite: $ 8x^3 + 27 = (2x)^3 + 3^3 $

Use sum of cubes:
$$
(2x)^3 + 3^3 = (2x + 3)((2x)^2 - (2x)(3) + 3^2) = (2x + 3)(4x^2 - 6x + 9)
$$

✔ Answer: $ (2x + 3)(4x^2 - 6x + 9) $

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5) $ a^3 - 216 $

$ 216 = 6^3 $, so $ a^3 - 6^3 $

Use difference of cubes:
$$
a^3 - 216 = (a - 6)(a^2 + 6a + 36)
$$

✔ Answer: $ (a - 6)(a^2 + 6a + 36) $

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6) $ 64x^3 - 27 $

$ 64x^3 = (4x)^3 $, $ 27 = 3^3 $

So: $ (4x)^3 - 3^3 $

Use difference of cubes:
$$
(4x - 3)((4x)^2 + (4x)(3) + 3^2) = (4x - 3)(16x^2 + 12x + 9)
$$

✔ Answer: $ (4x - 3)(16x^2 + 12x + 9) $

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7) $ 27m^3 - 125 $

$ 27m^3 = (3m)^3 $, $ 125 = 5^3 $

So: $ (3m)^3 - 5^3 $

Use difference of cubes:
$$
(3m - 5)((3m)^2 + (3m)(5) + 5^2) = (3m - 5)(9m^2 + 15m + 25)
$$

✔ Answer: $ (3m - 5)(9m^2 + 15m + 25) $

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8) $ x^3 - 64 $

$ 64 = 4^3 $, so $ x^3 - 4^3 $

Use difference of cubes:
$$
x^3 - 64 = (x - 4)(x^2 + 4x + 16)
$$

✔ Answer: $ (x - 4)(x^2 + 4x + 16) $

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9) $ 432 + 250m^3 $

First, rearrange: $ 250m^3 + 432 $

Look for common factors. Both terms divisible by 2:
$$
= 2(125m^3 + 216)
$$

Now check:
$ 125m^3 = (5m)^3 $, $ 216 = 6^3 $

So: $ (5m)^3 + 6^3 $ → Sum of cubes!

$$
= 2[(5m + 6)((5m)^2 - (5m)(6) + 6^2)] = 2(5m + 6)(25m^2 - 30m + 36)
$$

✔ Answer: $ 2(5m + 6)(25m^2 - 30m + 36) $

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10) $ 81x^3 + 192 $

Factor out GCF first:
Both 81 and 192 divisible by 3:
$$
= 3(27x^3 + 64)
$$

Now: $ 27x^3 = (3x)^3 $, $ 64 = 4^3 $

So: $ (3x)^3 + 4^3 $ → Sum of cubes!

$$
= 3[(3x + 4)((3x)^2 - (3x)(4) + 4^2)] = 3(3x + 4)(9x^2 - 12x + 16)
$$

✔ Answer: $ 3(3x + 4)(9x^2 - 12x + 16) $

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✔ Final Answers:

| Problem | Factored Form |
|--------|----------------|
| 1) $ x^3 + 8 $ | $ (x + 2)(x^2 - 2x + 4) $ |
| 2) $ a^3 + 64 $ | $ (a + 4)(a^2 - 4a + 16) $ |
| 3) $ a^3 + 216 $ | $ (a + 6)(a^2 - 6a + 36) $ |
| 4) $ 27 + 8x^3 $ | $ (2x + 3)(4x^2 - 6x + 9) $ |
| 5) $ a^3 - 216 $ | $ (a - 6)(a^2 + 6a + 36) $ |
| 6) $ 64x^3 - 27 $ | $ (4x - 3)(16x^2 + 12x + 9) $ |
| 7) $ 27m^3 - 125 $ | $ (3m - 5)(9m^2 + 15m + 25) $ |
| 8) $ x^3 - 64 $ | $ (x - 4)(x^2 + 4x + 16) $ |
| 9) $ 432 + 250m^3 $ | $ 2(5m + 6)(25m^2 - 30m + 36) $ |
| 10) $ 81x^3 + 192 $ | $ 3(3x + 4)(9x^2 - 12x + 16) $ |

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📝 Tips for Factoring Sum/Difference of Cubes:

1. Identify if the expression is a sum or difference of two perfect cubes.
2. Write each term as a cube: $ a^3 \pm b^3 $
3. Apply the correct formula.
4. Always check for GCF first (like in #9 and #10).
5. The trinomial part does not factor further over real numbers.

Let me know if you'd like a video explanation or practice quiz!