Question

Решите: 1)(a+b)^3 2)(a-1)^3 3)(b+1)^3 4)(c-m)^35)(m-d)^36)(d-2)^37)(3+d)^38)(5+a)^39)(2a+3b)10)(4b-a)^3​

Answer 1

$(a+b)^3=a^3+3a^2b+3ab^2+b^3=a^3+b^3+3ab\, (a+b)\\\\(a-1)^3=a^3-3a^2+3a-1\\\\(b+1)^3=b^3+3b^2+3b+1\\\\(c-m)^3=c^3-3c^2m+3cm^2-m^3\\\\(m-d)^3=m^3-3m^2d+3md^2-d^3\\\\(d-2)^3=d^3-6d^2+12d-8\\\\(3+d)^3=27+27d+9d^2+d^3\\\\(5+a)^3=125+75a+15a^2+a^3\\\\(2a+3b)^3=8a^3+36a^2b+54ab^2+27b^3\\\\(4b-a)^3=64b^3-48b^2a+12ba^2-a^3$

Answer 2

To expand these expressions, we can use the binomial expansion formula, which is:

(a + b)^n = Σ (n choose k) * a^(n-k) * b^k, where Σ is the summation symbol, n is the exponent, k ranges from 0 to n, and (n choose k) is the binomial coefficient, given by (n! / (k! * (n - k)!)).

Let's expand each expression:

1. (a + b)^3 = (3 choose 0) * a^3 * b^0 + (3 choose 1) * a^2 * b^1 + (3 choose 2) * a^1 * b^2 + (3 choose 3) * a^0 * b^3 = 1 * a^3 * b^0 + 3 * a^2 * b^1 + 3 * a^1 * b^2 + 1 * a^0 * b^3 = a^3 + 3a^2b + 3ab^2 + b^3

2. (a - 1)^3 = (3 choose 0) * a^3 * (-1)^0 + (3 choose 1) * a^2 * (-1)^1 + (3 choose 2) * a^1 * (-1)^2 + (3 choose 3) * a^0 * (-1)^3 = 1 * a^3 + 3 * a^2 * (-1) + 3 * a^1 * 1 + (-1)^3 = a^3 - 3a^2 + 3a - 1

3. (b + 1)^3 = (3 choose 0) * b^3 * 1^0 + (3 choose 1) * b^2 * 1^1 + (3 choose 2) * b^1 * 1^2 + (3 choose 3) * b^0 * 1^3 = 1 * b^3 + 3 * b^2 + 3 * b^1 + 1 * 1 = b^3 + 3b^2 + 3b + 1

4. (c - m)^3 Using the same formula, we get: = c^3 - 3c^2m + 3cm^2 - m^3

5. (m - d)^3 = m^3 - 3m^2d + 3md^2 - d^3

6. (d - 2)^3 = d^3 - 3d^2 * 2 + 3d * 2^2 - 2^3 = d^3 - 6d^2 + 12d - 8

7. (3 + d)^3 = 3^3 + 3 * 3^2 * d + 3 * 3 * d^2 + d^3 = 27 + 27d + 9d^2 + d^3

8. (5 + a)^3 = 5^3 + 3 * 5^2 * a + 3 * 5 * a^2 + a^3 = 125 + 75a + 15a^2 + a^3

9. (2a + 3b)^3 = (2a)^3 + 3 * (2a)^2 * 3b + 3 * 2a * (3b)^2 + (3b)^3 = 8a^3 + 36a^2b + 54ab^2 + 27b^3

10. (4b - a)^3 = (4b)^3 - 3 * (4b)^2 * a + 3 * 4b * a^2 - a^3 = 64b^3 - 48ab^2 + 12a^2b - a^3

Keep in mind that I've only expanded these expressions using the binomial expansion formula. If you need further simplifications or specific numerical evaluations, let me know!