Question

(2x-1)^2 (4+а)^2 (2а+3b)^2 (x^3-3)^3

Answer 1

(2x-1)² = 4х²-4х+1

(4+а)² = 16+8а+а²

(2а+3b)² = 4a²+12ab+9b²

(x³-3)³ = x⁹-9x⁶+27x-27

Answer 2

$1)(2x - 1 {)}^{2} = (2x {)}^{2} - 2 \times 2x \times 1 + {1}^{2} = \\ 4 {x}^{2} - 4x + 1 \\ 2)(4 + a {)}^{2} = {4}^{2} + 2 \times 4 \times a + {a}^{2} = \\ 16 + 8a + {a}^{2} \\ (2a + 3b {)}^{2} = (2a {)}^{2} + 2 \times 2a \times 3b + (3b {)}^{2} = \\ 4 {a}^{2} + 12ab + 9 {b}^{2} \\ ( {x}^{3} - 3 {)}^{3} = ( {x}^{3} {)}^{3} - 3 \times ( {x}^{3} {)}^{2} \times 3 - 3 \times {x}^{3} \times 9 - 27 = \\ {x}^{9} + 9 {x}^{6} - 27 {x}^{3} - 27$

Answer 3

To simplify the given expression, you can expand each binomial term and then combine like terms. Let's go step by step:

1. Expand (2x-1)^2: (2x-1)^2 = (2x-1)(2x-1) = 4x^2 - 2x - 2x + 1 = 4x^2 - 4x + 1

2. Expand (4+a)^2: (4+a)^2 = (4+a)(4+a) = 16 + 4a + 4a + a^2 = 16 + 8a + a^2

3. Expand (2a+3b)^2: (2a+3b)^2 = (2a+3b)(2a+3b) = 4a^2 + 6ab + 6ab + 9b^2 = 4a^2 + 12ab + 9b^2

4. Expand (x^3-3)^3: (x^3-3)^3 = (x^3-3)(x^3-3)(x^3-3) = (x^6 - 6x^3 + 9)(x^3 - 3) = x^9 - 3x^6 - 6x^6 + 18x^3 + 9x^3 - 27 = x^9 - 9x^6 + 27x^3 - 27

Now, let's multiply all the expanded expressions together:

(4x^2 - 4x + 1)(16 + 8a + a^2)(4a^2 + 12ab + 9b^2)(x^9 - 9x^6 + 27x^3 - 27)

To multiply them, you can use the distributive property and combine like terms. The result will be a long polynomial expression with terms involving different powers of x and a, as well as terms involving both x and a variables. However, the final expression will be too lengthy to write out here.

If you have a specific value for 'a' and 'b' or if there's a particular simplification you'd like to achieve, please provide those details, and I can help you further.