Question

(b-y)^3= (3+n)^3= (27+1)^3= (6m+3y)= (h-1)^3= P.S. ^3 ОЗНОЧАЕТ В КУБЕ ИЛИ СТЕПЕНЬ (НАПРИМЕР h в 3 степени

Answer 1

$(a+b)^3=a^3+ 3a^2b+3ab^2+ b^3=a^3+b^3+3ab(a+b)\\\\(a-b)^3=a^3-3a^2b+3ab^2-b^3=a^3-b^3-3ab(a-b)\\\\1)\; \; (b-y)^3=b^3-3b^2y+2by^2-y^3\\\\2)\; \; (3+n)^3=27+27n+9n^2+n^3\\\\3)\; \; (27+1)^3=28^3=21952\\\\4)\; \; (8m+3y)^3=(8m)^3+3\cdot (8m)^2\cdot 3y+3\cdot 8m\cdot (3y)^2+(3y)^3=\\\\=512\, m^3+57\, 6m^2y+216\, my^2+27n^3\\\\5)\; \; (h-1)^3=h^3-3h^2+3h-1$

Answer 2

1)b³-3b²y+3by²-y³
2)27+27n²+9n²+n³
3)28³
4)Если там также должен быть куб(его там нету, тогда)
216m³+324m²+162my²+27y³
5)h³-3h²+3h-1

Answer 3

It seems that you have provided a set of equations where each equation involves an expression raised to the power of 3. Here's how you can solve each of them:

1. (b - y)^3:

To expand this expression, you can use the binomial formula, which states that:

(a + b)^n = ∑(k=0 to n) [n choose k] * a^(n-k) * b^k

where [n choose k] is the binomial coefficient.

Using this formula with a = b and n = 3, we get:

(b - y)^3 = ∑(k=0 to 3) [3 choose k] * b^(3-k) * (-y)^k

Simplifying this expression, we get:

(b - y)^3 = b^3 - 3b^2y + 3by^2 - y^3

2. (3 + n)^3:

Using the binomial formula again with a = 3 and n = 3, we get:

(3 + n)^3 = ∑(k=0 to 3) [3 choose k] * 3^(3-k) * n^k

Simplifying this expression, we get:

(3 + n)^3 = 27 + 27n + 9n^2 + n^3

3. (27 + 1)^3:

This expression can be simplified using the identity:

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

where a = 27 and b = 1. Substituting these values, we get:

(27 + 1)^3 = 27^3 + 3(27^2)(1) + 3(27)(1^2) + 1^3

Simplifying this expression, we get:

(27 + 1)^3 = 19683

4. (6m + 3y):

This expression can be factored by taking out the common factor of 3:

6m + 3y = 3(2m + y)

5. (h - 1)^3:

Using the binomial formula with a = h and b = -1 and n = 3, we get:

(h - 1)^3 = ∑(k=0 to 3) [3 choose k] * h^(3-k) * (-1)^k

Simplifying this expression, we get:

(h - 1)^3 = h^3 - 3h^2 + 3h - 1

I hope this helps you solve your equations! Let me know if you have any further questions.