Factoring Sum and Difference of Cubes worksheet with problems and a motivational quote.

It looks like you've uploaded an image of a worksheet titled "Factoring Sum and Difference of Cubes" with several algebraic expressions to factor. Since I can't view images directly, I’ll solve the problems based on the text you provided in your message.

Here are the problems listed:

1. $ 375y^3 - 81 $
2. $ 20y^3 + 16 $
3. $ 2y^3 - 128 $
4. $ 64 - 27y^3 $
5. $ 8y^3 = 125 $ (Note: This seems to be an equation, not an expression to factor — possibly a typo)
6. $ 250y^3 + 128 $
7. $ 27y^3 + 125 $
8. $ 8y^3 = 125 $ (Again, this is an equation — likely meant to be factored or solved)
9. $ 500y^3 - 32 $
10. $ 80y^3 + 27 $
11. $ 1575y^3 - 216 $
12. $ 36y^3 + 20y^3 $

Let’s go through each one carefully and apply the sum and difference of cubes formulas:

🔷 Sum of Cubes:

$$
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
$$

🔷 Difference of Cubes:

$$
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
$$

We’ll factor each expression step by step.

---

1. $ 375y^3 - 81 $

Factor out the GCF first:
- $ 375y^3 - 81 = 3(125y^3 - 27) $
- Now: $ 125y^3 = (5y)^3 $, $ 27 = 3^3 $
- So: $ 125y^3 - 27 = (5y)^3 - 3^3 $
- Use difference of cubes:
$$
= (5y - 3)((5y)^2 + (5y)(3) + 3^2) = (5y - 3)(25y^2 + 15y + 9)
$$
So full factorization:
$$
\boxed{3(5y - 3)(25y^2 + 15y + 9)}
$$

---

2. $ 20y^3 + 16 $

Factor out GCF:
- $ 20y^3 + 16 = 4(5y^3 + 4) $
- $ 5y^3 + 4 $: Not a perfect cube → cannot factor using sum/difference of cubes
- So only factorable as:
$$
\boxed{4(5y^3 + 4)}
$$

(Note: This is not a sum of cubes because $ 5y^3 $ and $ 4 $ are not both perfect cubes.)

---

3. $ 2y^3 - 128 $

Factor out GCF:
- $ 2(y^3 - 64) $
- $ y^3 - 64 = y^3 - 4^3 $ → difference of cubes
- $ = (y - 4)(y^2 + 4y + 16) $
So:
$$
\boxed{2(y - 4)(y^2 + 4y + 16)}
$$

---

4. $ 64 - 27y^3 $

Rewrite: $ 64 - 27y^3 = 4^3 - (3y)^3 $
→ difference of cubes
- $ = (4 - 3y)(4^2 + 4(3y) + (3y)^2) $
- $ = (4 - 3y)(16 + 12y + 9y^2) $
So:
$$
\boxed{(4 - 3y)(16 + 12y + 9y^2)}
$$

---

5. $ 8y^3 = 125 $

This is an equation, not an expression to factor.
Solve:
- $ 8y^3 = 125 $
- $ y^3 = \frac{125}{8} $
- $ y = \sqrt[3]{\frac{125}{8}} = \frac{5}{2} $
So solution: $ \boxed{y = \frac{5}{2}} $

But if it was meant to be factored: $ 8y^3 - 125 = (2y)^3 - 5^3 $
→ $ = (2y - 5)(4y^2 + 10y + 25) $

Assuming it was a typo and should be $ 8y^3 - 125 $, then:
$$
\boxed{(2y - 5)(4y^2 + 10y + 25)}
$$

---

6. $ 250y^3 + 128 $

Factor out GCF:
- $ 250y^3 + 128 = 2(125y^3 + 64) $
- $ 125y^3 = (5y)^3 $, $ 64 = 4^3 $
- So: $ (5y)^3 + 4^3 $ → sum of cubes
- $ = (5y + 4)((5y)^2 - (5y)(4) + 4^2) = (5y + 4)(25y^2 - 20y + 16) $
So:
$$
\boxed{2(5y + 4)(25y^2 - 20y + 16)}
$$

---

7. $ 27y^3 + 125 $

$ 27y^3 = (3y)^3 $, $ 125 = 5^3 $
→ sum of cubes
- $ = (3y + 5)((3y)^2 - (3y)(5) + 5^2) $
- $ = (3y + 5)(9y^2 - 15y + 25) $
$$
\boxed{(3y + 5)(9y^2 - 15y + 25)}
$$

---

8. $ 8y^3 = 125 $

Same as #5 → likely meant to be $ 8y^3 - 125 $
As above:
$$
\boxed{(2y - 5)(4y^2 + 10y + 25)}
$$

---

9. $ 500y^3 - 32 $

Factor out GCF:
- $ 500y^3 - 32 = 4(125y^3 - 8) $
- $ 125y^3 = (5y)^3 $, $ 8 = 2^3 $
- $ = (5y)^3 - 2^3 $ → difference of cubes
- $ = (5y - 2)(25y^2 + 10y + 4) $
So:
$$
\boxed{4(5y - 2)(25y^2 + 10y + 4)}
$$

---

10. $ 80y^3 + 27 $

Factor out GCF? 80 and 27 have no common factor.
Check if it's a sum of cubes:
- $ 80y^3 = ? $ → $ 80 $ is not a perfect cube
- $ 27 = 3^3 $, but $ 80y^3 $ is not a perfect cube
→ Not a sum of cubes
But try factoring:
Wait: $ 80y^3 + 27 $ — not a perfect cube pair.

Wait: Let’s check if we can write as:
No — not a sum of cubes.

But maybe factor as:
- $ 80y^3 + 27 $ — no common factors, not a sum of cubes.
So perhaps it’s not factorable via sum/difference of cubes.

But wait: Is there a typo?

Alternatively, could it be $ 8y^3 + 27 $? That would be $ (2y)^3 + 3^3 $, which is a sum of cubes.

But as written: $ 80y^3 + 27 $ → not factorable via sum/difference of cubes.

So unless there's a typo, answer is:
$$
\boxed{80y^3 + 27} \quad \text{(not factorable using sum/difference of cubes)}
$$

But let’s double-check: Maybe factor out something?

GCF of 80 and 27 is 1 → cannot factor.

So no, not factorable using these methods.

---

11. $ 1575y^3 - 216 $

Factor out GCF:
- $ 1575 $ and $ 216 $
- Find GCF:
$ 1575 = 25 × 63 = 5² × 3² × 7 $
$ 216 = 6³ = 2³ × 3³ $
Common factor: $ 3² = 9 $

So:
- $ 1575y^3 - 216 = 9(175y^3 - 24) $

Now check $ 175y^3 - 24 $:
- $ 175 = 25×7 $, $ 24 = 8×3 $ → neither is a perfect cube

Not a difference of cubes.

Wait: Try factoring differently?

Wait — original: $ 1575y^3 - 216 $

Try factoring $ 1575y^3 - 216 $ directly.

Check if either term is a perfect cube:
- $ 216 = 6^3 $ → yes!
- $ 1575y^3 = ? $ → $ 1575 = 25 × 63 $, not a perfect cube

But $ 1575y^3 = ( \sqrt[3]{1575} y )^3 $ → not rational

So unless $ 1575y^3 $ is a perfect cube, we can’t use sum/difference of cubes.

But $ 1575 = 25 × 63 = 5² × 3² × 7 $ → not a perfect cube

So not a difference of cubes.

But wait: Could it be $ 1575y^3 - 216 = ? $

Wait: $ 216 = 6^3 $, but $ 1575 $ is not a cube.

So unless it's $ 1575y^3 - 216 $, and we can't factor it as a difference of cubes.

But perhaps there’s a typo — maybe it's $ 1575y^3 - 216 $, but we can still try:

Alternatively, factor out 9:
- $ 1575y^3 - 216 = 9(175y^3 - 24) $

Still not helpful.

Wait — maybe $ 1575y^3 = 27 × 58.33 $? No.

Wait — what if it's $ 1575y^3 - 216 = ? $

Let me see: $ 1575 = 25 × 63 $, $ 216 = 6^3 $

No.

Wait — perhaps the problem is $ 1575y^3 - 216 $, and we need to factor it differently.

Alternatively, could it be $ 1575y^3 - 216 = ? $

Wait — maybe it's supposed to be $ 1575y^3 - 216 = 9(175y^3 - 24) $, but still not a cube.

Wait — another idea: maybe it's $ 1575y^3 - 216 = ? $

Actually, let's try to see if $ 1575y^3 - 216 $ is divisible by $ y^3 - a $, but that’s complicated.

Alternatively, perhaps it's a typo and meant to be $ 1575y^3 - 216 $, but without being a perfect cube, we can’t use sum/difference of cubes.

So unless we made a mistake, this expression does not factor nicely using sum/difference of cubes.

But let’s check: Is $ 1575y^3 $ close to a cube?

$ 1575 ≈ 1575 $, $ 11^3 = 1331 $, $ 12^3 = 1728 $ → so no.

So conclusion: Not factorable via sum/difference of cubes

But wait — maybe we missed something.

Wait — look at $ 1575y^3 - 216 $. Factor out 9:
- $ 1575 ÷ 9 = 175 $, $ 216 ÷ 9 = 24 $
- So $ 9(175y^3 - 24) $

Still not a cube.

So unless the problem has a typo, it cannot be factored using sum/difference of cubes.

But let’s suppose it was $ 1575y^3 - 216 $, and we leave it as is.

Alternatively, maybe it's $ 1575y^3 - 216 $, but no.

Wait — perhaps it's $ 1575y^3 - 216 = 9(175y^3 - 24) $, and that's the best we can do.

So:
$$
\boxed{9(175y^3 - 24)} \quad \text{(cannot be factored further using sum/difference of cubes)}
$$

---

12. $ 36y^3 + 20y^3 $

This is just addition:
- $ 36y^3 + 20y^3 = 56y^3 $
So:
$$
\boxed{56y^3}
$$

But if it was meant to be $ 36y^3 + 20y^3 $, then yes, simplify.

But if it was $ 36y^3 + 20y^3 $, it’s not a sum of cubes.

Wait — maybe it was meant to be $ 36y^3 + 20y^3 $, but that’s just combining like terms.

So final answer:
$$
\boxed{56y^3}
$$

---

## ✔ Final Answers Summary:

| Problem | Factored Form |
|--------|----------------|
| 1 | $ 3(5y - 3)(25y^2 + 15y + 9) $ |
| 2 | $ 4(5y^3 + 4) $ (cannot factor further) |
| 3 | $ 2(y - 4)(y^2 + 4y + 16) $ |
| 4 | $ (4 - 3y)(16 + 12y + 9y^2) $ |
| 5 | Likely typo: $ 8y^3 - 125 = (2y - 5)(4y^2 + 10y + 25) $ |
| 6 | $ 2(5y + 4)(25y^2 - 20y + 16) $ |
| 7 | $ (3y + 5)(9y^2 - 15y + 25) $ |
| 8 | Likely typo: $ 8y^3 - 125 = (2y - 5)(4y^2 + 10y + 25) $ |
| 9 | $ 4(5y - 2)(25y^2 + 10y + 4) $ |
| 10 | $ 80y^3 + 27 $ — not factorable using sum/difference of cubes |
| 11 | $ 1575y^3 - 216 = 9(175y^3 - 24) $ — not factorable |
| 12 | $ 56y^3 $ |

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If you can confirm whether problems 5, 8, 10, 11 were intended to be equations or expressions, or if there are typos (e.g., $ 8y^3 + 27 $ instead of $ 80y^3 + 27 $), I can refine the answers.

But based on current input, here are the solutions.