Question

Use the binomial theorem to expand (d-4b)^3

Answer 1

Answer: 12bd^2+48b^2d-64b^3[/tex]

Explanation:

we have general formula  for binomial expansion which is

$(a+b)^n=a^n+na^{n-1} b +\frac{n(n-1)}{2} a^{n-2}b^2+----------+b^n$

Since, we have to evaluate

$(d-4b)^3$

here a=d and b=$-4band n=3$

Substituting values in the formula we will get

$d^3+3d^{3-1}(-4b)+ \frac{(3)(3-1)}{2} d^{3-2}(-4b)^2+(-4b)^3 \\\\d^3-12bd^2+48b^2d-64b^3$

Answer 2

The binomial expansion of the expression ${\left({d - 4b} \right)^3}$ is $\boxed{{{\mathbf{d}}^{\mathbf{3}}}{\mathbf{ - 12b}}{{\mathbf{d}}^{\mathbf{2}}}{\mathbf{ + 48}}{{\mathbf{b}}^{\mathbf{2}}}{\mathbf{d - 64}}{{\mathbf{b}}^{\mathbf{3}}}}$.

Further explanation:

It is given that the expression is ${\left({d - 4b}\right)^3}$.

Now, the expansion of the expression is simplified using binomial theorem.

The binomial is a polynomial with only two terms. The binomial theorem is used to simplify the algebraic expansion of any power of a binomial expression.

Now, consider $x$ and $y$ as the numbers and the binomial expression becomes $x + y$. If the power is assumed to be $n$  , then the binomial expression is ${\left({x + y}\right)^n}$.

The binomial expression ${\left({x + y}\right)^n}$ is expanded as follows:

For any positive integer $n$,the expanded form is given below.

${\left( {x + y}\right)^n} = {x^n} + \left(\begin{aligned}n\hfill\\1\hfill\\\end{gathered}\right){x^{n - 1}}y + \left(\begin{aligned}n\hfill\\2\hfill\\\end{aligned}\right){x^{n - 2}}{y^2}+\cdots+\left( \begin{aligned}n\hfill\\r \hfill\\\end{gathered}\right){x^{n - r}}{y^r}+\cdots + \left(\begin{aligned}\,\,\,n\hfill\\n - 1\hfill\\\end{aligned}\right)x{y^{n - 1}} + {y^n}$

For non-negative integers $n$ and $r$

with $r\leqslant n$, the expression is $\left(\begin{aligned}n\hfill\\r\hfill\\\end{aligned}\right) = \dfrac{{n!}}{{r!\left({n - r} \right)!}}$ where, $n! = n\cdot \left({n - 1}\right)\cdot\left( {n - 2}\right)\cdots 3\cdot 2\cdot 1$.

The expression ${\left({d - 4b}\right)^3}$ is expanded using the binomial theorem as follows:

$\begin{aligned}(d-4b)^{3}&=\left[\dbinom{3}{0}(d)^{3}(-4b)^{0}+\dbinom{3}{1}(d)^{2}(-4b)^{1}+\dbinom{3}{2}(d)^{1}(-4b)^{2}+\dbinom{3}{3}(d)^{0}(-4b)^{3}+\right]\\&=\left[\dfrac{3!}{3!\cdot 0!}d^{3}(1)+\dfrac{3!}{1!\cdot 2!}(-d^{2}4b)+\dfrac{3!}{2!\cdot 1!}d(16b^{2})+\dfrac{3!}{3!\cdot 0!}d(1)(-64b^{3})\right]\end{aligned}$

Simplify the above equation as follows:

$\begin{aligned}{\left({d - 4b}\right)^3}&=\left[{{d^3} + 3\left({ - {d^2}4b} \right) + 3d\left( {16{b^2}}\right)+\left({ - 64{b^3}}\right)} \right]\\&={d^3} - 12b{d^2} + 48{b^2}d - 64{b^3}\\\end{aligned}$

Thus, the binomial expansion of the expression ${\left( {d - 4b}\right)^3}$ is $\boxed{{{\mathbf{d}}^{\mathbf{3}}}{\mathbf{ - 12b}}{{\mathbf{d}}^{\mathbf{2}}}{\mathbf{ + 48}}{{\mathbf{b}}^{\mathbf{2}}}{\mathbf{d - 64}}{{\mathbf{b}}^{\mathbf{3}}}}$.

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Answer Details:

Grade: Junior High School

Subject: Mathematics

Chapter: Binomial Theorem

Keywords: Binomial theorem, linear equation, system of linear equations in two variables,${\left({d - 4b}\right)^3}$ , expression, theorem, expansion