Question

(-5a+b)^3 Решить по формуле.

Answer 1

Использовать переместительный закон, чтобы изменить порядок членов.

(b-5a)^3

Используя формулу

(x-y)^3=x^3-3x^2y+3xy^2-y^3 записать выражение в развёрнутом виде

b^3-15ab^2+75a^2b-125a^3

Answer 2

To solve the expression $(-5a + b)^3$ using the binomial formula, we can expand it step by step. The binomial formula, also known as the binomial theorem, is given by:

$(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$

where $n$ is the power, $\binom{n}{k}$ is the binomial coefficient (also known as "n choose k"), and $x$ and $y$ are the terms of the binomial expression.

In our case, $x = -5a$ and $y = b$, and $n = 3$. So, let's expand $(-5a + b)^3$ step by step:

Step 1: Find the coefficients for each term using the binomial coefficient formula $\binom{n}{k}$: $\binom{3}{0} = 1$ $\binom{3}{1} = 3$ $\binom{3}{2} = 3$ $\binom{3}{3} = 1$

Step 2: Write out the expanded form using the binomial formula: $(x + y)^3 = \binom{3}{0} x^{3-0} y^0 + \binom{3}{1} x^{3-1} y^1 + \binom{3}{2} x^{3-2} y^2 + \binom{3}{3} x^{3-3} y^3$

Step 3: Substitute $x = -5a$ and $y = b$ into the formula: $( -5a + b )^3 = \binom{3}{0} (-5a)^3 b^0 + \binom{3}{1} (-5a)^2 b^1 + \binom{3}{2} (-5a)^1 b^2 + \binom{3}{3} (-5a)^0 b^3$

Step 4: Simplify each term: $\binom{3}{0} (-5a)^3 b^0 = 1 \cdot (-5a)^3 \cdot 1 = -125a^3$ $\binom{3}{1} (-5a)^2 b^1 = 3 \cdot (-5a)^2 \cdot b = 75a^2b$ $\binom{3}{2} (-5a)^1 b^2 = 3 \cdot (-5a)^1 \cdot b^2 = -15ab^2$ $\binom{3}{3} (-5a)^0 b^3 = 1 \cdot b^3 = b^3$

Step 5: Combine the simplified terms: $(-5a + b)^3 = -125a^3 + 75a^2b - 15ab^2 + b^3$