Question

(3-x)^6 разложение бинома

Answer 1

Ответ:

x^ 6 - 18x^5 + 135x^4 - 540x^3 + 1215x^2 - 1458x + 729

Пошаговое объяснение:

Answer 2

To expand the binomial $(3-x)^6$, we can use the binomial theorem, which states that for any real numbers $a$ and $b$ and a non-negative integer $n$, the expansion of $(a+b)^n$ is given by:

$(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

where $\binom{n}{k}$ represents the binomial coefficient, which is the number of ways to choose $k$ elements from a set of $n$ elements and is defined as:

$\binom{n}{k} = \frac{n!}{k!(n-k)!}$

In our case, $a = 3$ and $b = -x$, and $n = 6$. Let's plug these values into the formula to expand $(3-x)^6$:

$(3-x)^6 = \binom{6}{0} 3^6 (-x)^0 + \binom{6}{1} 3^5 (-x)^1 + \binom{6}{2} 3^4 (-x)^2 + \binom{6}{3} 3^3 (-x)^3 + \binom{6}{4} 3^2 (-x)^4 + \binom{6}{5} 3^1 (-x)^5 + \binom{6}{6} 3^0 (-x)^6$

Now, let's calculate each term:

$\binom{6}{0} 3^6 (-x)^0 = 1 \cdot 3^6 \cdot 1 = 729$ $\binom{6}{1} 3^5 (-x)^1 = 6 \cdot 3^5 \cdot (-x) = -1458x$ $\binom{6}{2} 3^4 (-x)^2 = 15 \cdot 3^4 \cdot x^2 = 405x^2$ $\binom{6}{3} 3^3 (-x)^3 = 20 \cdot 3^3 \cdot (-x)^3 = -540x^3$ $\binom{6}{4} 3^2 (-x)^4 = 15 \cdot 3^2 \cdot x^4 = 405x^4$ $\binom{6}{5} 3^1 (-x)^5 = 6 \cdot 3^1 \cdot (-x)^5 = -162x^5$ $\binom{6}{6} 3^0 (-x)^6 = 1 \cdot 3^0 \cdot x^6 = x^6$