Evaluate algebraic expressions with given values a = -3 and b = -1.

We are given:
a = -3, b = -1

We will evaluate each expression one by one, carefully substituting the values and following order of operations (PEMDAS: Parentheses, Exponents, Multiplication/Division, Addition/Subtraction).

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Let’s go step by step.

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① $ a^2 $

$ (-3)^2 = 9 $

② $ 2b^2 $

First: $ b^2 = (-1)^2 = 1 $
Then: $ 2 \times 1 = 2 $

③ $ b^3 $

$ (-1)^3 = -1 $

④ $ (-b)^3 $

$ -b = -(-1) = 1 $ → $ 1^3 = 1 $

⑤ $ a + b $

$ -3 + (-1) = -4 $

⑥ $ a - b $

$ -3 - (-1) = -3 + 1 = -2 $

⑦ $ -a - b $

$ -(-3) - (-1) = 3 + 1 = 4 $

⑧ $ ab $

$ (-3)(-1) = 3 $

⑨ $ a^2 + b^2 $

$ 9 + 1 = 10 $

Ⓐ $ a^2 - b^2 $

$ 9 - 1 = 8 $

Ⓑ $ a^3 + b^3 $

$ a^3 = (-3)^3 = -27 $, $ b^3 = (-1)^3 = -1 $
→ $ -27 + (-1) = -28 $

Ⓒ $ a^3 - b^3 $

$ -27 - (-1) = -27 + 1 = -26 $

Ⓓ $ 2a^2 + b $

$ 2*(9) + (-1) = 18 - 1 = 17 $

Ⓔ $ 2a^2 - b $

$ 18 - (-1) = 18 + 1 = 19 $

Ⓕ $ 2a^2 - 2b $

$ 18 - 2*(-1) = 18 + 2 = 20 $

Ⓖ $ 2a^2 - 2b^2 $

$ 18 - 2*(1) = 18 - 2 = 16 $

Ⓗ $ 2a^2 - 2b^3 $

$ 18 - 2*(-1) = 18 + 2 = 20 $

Ⓘ $ 2b^3 - 2a^3 $

$ 2*(-1) - 2*(-27) = -2 + 54 = 52 $

Ⓙ $ 2(a + b) $

$ 2*(-4) = -8 $

Ⓚ $ 2(a - b) $

$ 2*(-2) = -4 $

Ⓛ $ 3(b - a) $

$ b - a = -1 - (-3) = -1 + 3 = 2 $ → $ 3*2 = 6 $

$ a(a - b) $

$ a - b = -2 $ → $ (-3)*(-2) = 6 $

$ b(b - a) $

$ b - a = 2 $ → $ (-1)*2 = -2 $

Ⓞ $ b(a - 2b) $

First: $ 2b = 2*(-1) = -2 $
Then: $ a - 2b = -3 - (-2) = -3 + 2 = -1 $
Then: $ b * (-1) = (-1)*(-1) = 1 $

Ⓟ $ a^2(a + b^2) $

$ a^2 = 9 $, $ b^2 = 1 $ → $ a + b^2 = -3 + 1 = -2 $
→ $ 9 * (-2) = -18 $

Ⓠ $ b^2(a^2 - b) $

$ b^2 = 1 $, $ a^2 = 9 $, so $ 9 - (-1) = 10 $
→ $ 1 * 10 = 10 $

$ (a + b)^2 $

$ a + b = -4 $ → $ (-4)^2 = 16 $

$ (a - b)^2 $

$ a - b = -2 $ → $ (-2)^2 = 4 $

Ⓣ $ (2a - b)^2 $

$ 2a = 2*(-3) = -6 $, then $ -6 - (-1) = -6 + 1 = -5 $
→ $ (-5)^2 = 25 $

Ⓤ $ (a - b)^3 $

$ a - b = -2 $ → $ (-2)^3 = -8 $

$ a(a + b)^2 $

$ a + b = -4 $, squared is 16 → $ (-3)*16 = -48 $

Ⓦ $ (2a^2 - 2b)^2 $

From earlier: $ 2a^2 = 18 $, $ 2b = -2 $ → $ 18 - (-2) = 20 $? Wait — no:

Wait! Expression is $ (2a^2 - 2b)^2 $

We have:
$ 2a^2 = 18 $
$ 2b = 2*(-1) = -2 $
So: $ 18 - (-2) = 18 + 2 = 20 $? No — wait: it's minus 2b, not minus negative.

Actually:
Expression: $ 2a^2 - 2b $ → that’s what we did in Ⓕ → which was 20? Let me check:

In Ⓕ: $ 2a^2 - 2b = 18 - 2*(-1) = 18 + 2 = 20 $. Yes.

But here it’s $ (2a^2 - 2b)^2 $ → so same as Ⓕ but squared → $ 20^2 = 400 $

Wait — let me double-check the expression: Ⓦ is $ (2a^2 - 2b)^2 $

Yes, so inside: $ 2a^2 - 2b = 18 - 2*(-1) = 18 + 2 = 20 $ → square is 400.

BUT — hold on! Is it possible I misread? Let me look again at original image description.

Actually, looking back: Ⓦ is written as $ (2a^2 - 2b)^2 $ — yes.

But let me recalculate to be safe:

$ 2a^2 = 2*(9) = 18 $
$ 2b = 2*(-1) = -2 $
So $ 2a^2 - 2b = 18 - (-2) = 20 $? No — subtraction of 2b means: if b is -1, then 2b is -2, so subtracting 2b is subtracting (-2), which is adding 2.

Yes: $ 18 - (-2) = 20 $. So squared is 400.

Okay.

Ⓧ $ \frac{a}{b} $

$ \frac{-3}{-1} = 3 $

$ \frac{a + b}{b} $

Numerator: $ -3 + (-1) = -4 $, denominator: -1 → $ \frac{-4}{-1} = 4 $

$ \frac{a - b}{b} $

Numerator: $ -3 - (-1) = -2 $, denominator: -1 → $ \frac{-2}{-1} = 2 $

α $ \frac{a + b}{a - b} $

Numerator: -4, Denominator: -2 → $ \frac{-4}{-2} = 2 $

β $ \frac{a^2 + b}{a - b} $

Numerator: $ 9 + (-1) = 8 $, Denominator: -2 → $ \frac{8}{-2} = -4 $

γ $ \frac{a^2 + b}{(a - b)^2} $

Numerator: 8, Denominator: $ (-2)^2 = 4 $ → $ \frac{8}{4} = 2 $

Δ $ \frac{2(a + b)}{2(a + b)^2} $

Note: This simplifies algebraically to $ \frac{1}{a + b} $, as long as $ a + b ≠ 0 $, which it isn’t (-4).

But let’s compute numerically:

Numerator: $ 2*(-4) = -8 $
Denominator: $ 2*(-4)^2 = 2*16 = 32 $
→ $ \frac{-8}{32} = -\frac{1}{4} $

Alternatively, cancel 2: $ \frac{a + b}{(a + b)^2} = \frac{1}{a + b} = \frac{1}{-4} = -0.25 $

Same result.

ε $ \frac{2(b - a)}{2(a - b)^2} $

Simplify: Cancel 2 → $ \frac{b - a}{(a - b)^2} $

Note: $ b - a = -(a - b) $, so numerator = $ -(a - b) $, denominator = $ (a - b)^2 $

So overall: $ \frac{-(a - b)}{(a - b)^2} = \frac{-1}{a - b} $

Now plug in: $ a - b = -2 $ → $ \frac{-1}{-2} = \frac{1}{2} $

Numerically:

Numerator: $ 2*(b - a) = 2*(-1 - (-3)) = 2*(2) = 4 $
Denominator: $ 2*(a - b)^2 = 2*(-2)^2 = 2*4 = 8 $
→ $ \frac{4}{8} = \frac{1}{2} $

Perfect.

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Now, compiling all answers in order:

① 9
② 2
③ -1
④ 1
⑤ -4
⑥ -2
⑦ 4
⑧ 3
⑨ 10
Ⓐ 8
Ⓑ -28
-26
Ⓓ 17
Ⓔ 19
Ⓕ 20
Ⓖ 16
Ⓗ 20
Ⓘ 52
Ⓙ -8
Ⓚ -4
6
Ⓜ 6
Ⓝ -2
1
Ⓟ -18
Ⓠ 10
Ⓡ 16
Ⓢ 4
25
Ⓤ -8
Ⓥ -48
400
Ⓧ 3
Ⓨ 4
2
α 2
β -4
γ 2
Δ -1/4 or -0.25
ε 1/2 or 0.5

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Final Answer:

① 9
② 2
-1
④ 1
⑤ -4
⑥ -2
⑦ 4
⑧ 3
⑨ 10
Ⓐ 8
Ⓑ -28
Ⓒ -26
Ⓓ 17
Ⓔ 19
Ⓕ 20
Ⓖ 16
Ⓗ 20
Ⓘ 52
Ⓙ -8
Ⓚ -4
Ⓛ 6
Ⓜ 6
Ⓝ -2
Ⓞ 1
Ⓟ -18
Ⓠ 10
Ⓡ 16
4
Ⓣ 25
Ⓤ -8
Ⓥ -48
Ⓦ 400
3
Ⓨ 4
2
α 2
β -4
γ 2
Δ -1/4
ε 1/2