Question

How do you expand the binomial ${{\left( 2x-{{y}^{3}} \right)}^{7}}$?

Answer 1

From the given question we have to expand the binomial ${{\left( 2x-{{y}^{3}} \right)}^{7}}$. To expand this, we have to use binomial theorem i.e., the expansion of ${{\left( a+b \right)}^{n}}=\sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}.\left( {{a}^{n-k}}{{b}^{k}} \right)}$. Here we have to substitute 2x in place of $a$ and $-{{y}^{3}}$ in place of b. By using this binomial equation formula and after the simplification according to the formula we can expand the above binomial ${{\left( 2x-{{y}^{3}} \right)}^{7}}$.

Complete step by step solution:
From the given question we have to expand the binomial
As we know that we have to expand this by using binomial theorem. Binomial theorem helps in expanding the algebraic expansion of powers of a binomial into simplified form. According to the theorem, we can expand any polynomial ${{\left( a+b \right)}^{n}}$ into a sum involving terms of the form $c{{a}^{x}}{{b}^{y}}$.
Here the exponents x and y are nonnegative integers which obey the condition $x+y=n$. The coefficient c in the term of $c{{a}^{x}}{{b}^{y}}$ is known as the binomial coefficient.
Now, by using binomial theorem formula we have to expand the binomial ${{\left( 2x-{{y}^{3}} \right)}^{7}}$ as follows.
$\Rightarrow {{\left( 2x-{{y}^{3}} \right)}^{7}}={{\sum\limits_{k=0}^{7}{\dfrac{7!}{\left( 7-k \right)!k!}.\left( 2x \right)}}^{7-k}}.{{\left( -{{y}^{3}} \right)}^{k}}$
Now, we have to expand the above summation. So, the equation will be simplified as follows.

& \Rightarrow {{\left( 2x-{{y}^{3}} \right)}^{7}}=\dfrac{7!}{\left( 7-0 \right)!0!}.{{\left( 2x \right)}^{7-0}}.{{\left( -{{y}^{3}} \right)}^{0}}+\dfrac{7!}{\left( 7-1 \right)!1!}.{{\left( 2x \right)}^{7-1}}.{{\left( -{{y}^{3}} \right)}^{1}}+\dfrac{7!}{\left( 7-2 \right)!2!}.{{\left( 2x \right)}^{7-2}}.{{\left( -{{y}^{3}} \right)}^{2}} \\\ & +\dfrac{7!}{\left( 7-3 \right)!3!}.{{\left( 2x \right)}^{7-3}}.{{\left( -{{y}^{3}} \right)}^{3}}+\dfrac{7!}{\left( 7-4 \right)!4!}.{{\left( 2x \right)}^{7-4}}.{{\left( -{{y}^{3}} \right)}^{4}}+\dfrac{7!}{\left( 7-5 \right)!5!}.{{\left( 2x \right)}^{7-5}}.{{\left( -{{y}^{3}} \right)}^{5}} \\\ & +\dfrac{7!}{\left( 7-6 \right)!6!}.{{\left( 2x \right)}^{7-6}}.{{\left( -{{y}^{3}} \right)}^{6}}+\dfrac{7!}{\left( 7-7 \right)!7!}.{{\left( 2x \right)}^{7-7}}.{{\left( -{{y}^{3}} \right)}^{7}} \\\ \end{aligned}$$ Now, we have to simplify the above form. $$\begin{aligned} & \Rightarrow {{\left( 2x-{{y}^{3}} \right)}^{7}}=\left( 1.{{\left( -{{y}^{3}} \right)}^{0}}.{{\left( 2x \right)}^{7}} \right)+\left( 7.{{\left( -{{y}^{3}} \right)}^{1}}.{{\left( 2x \right)}^{6}} \right)+\left( 21.{{\left( -{{y}^{3}} \right)}^{2}}.{{\left( 2x \right)}^{5}} \right)+\left( 35.{{\left( -{{y}^{3}} \right)}^{3}}.{{\left( 2x \right)}^{4}} \right) \\\ & +\left( 35.{{\left( -{{y}^{3}} \right)}^{4}}.{{\left( 2x \right)}^{3}} \right)+\left( 21.{{\left( -{{y}^{3}} \right)}^{5}}.{{\left( 2x \right)}^{2}} \right)+\left( 7.{{\left( -{{y}^{3}} \right)}^{6}}.{{\left( 2x \right)}^{1}} \right)+\left( 1.{{\left( -{{y}^{3}} \right)}^{7}}.{{\left( 2x \right)}^{0}} \right) \\\ \end{aligned}$$ After the simplification the above binomial expression we get, $$\Rightarrow {{\left( 2x-{{y}^{3}} \right)}^{7}}=128{{x}^{7}}-448{{x}^{6}}{{y}^{3}}+672{{x}^{5}}{{y}^{6}}-560{{x}^{4}}{{y}^{9}}+280{{x}^{3}}{{y}^{12}}-84{{x}^{2}}{{y}^{15}}+14x{{y}^{18}}-{{y}^{21}}$$ **Therefore, for the given question the solution will be $${{\left( 2x-{{y}^{3}} \right)}^{7}}=128{{x}^{7}}-448{{x}^{6}}{{y}^{3}}+672{{x}^{5}}{{y}^{6}}-560{{x}^{4}}{{y}^{9}}+280{{x}^{3}}{{y}^{12}}-84{{x}^{2}}{{y}^{15}}+14x{{y}^{18}}-{{y}^{21}}$$.** **Note:** Students should know the expansions and binomial theorem. Student should be careful with signs and calculation. Students must have good knowledge in binomial theorems formula for example $${{\left( a+b \right)}^{n}}=\sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}.\left( {{a}^{n-k}}{{b}^{k}} \right)}$$. Students must be very careful in simplifying the question using the binomial formula we should not do mistake like, for example in this $$ {{\left( 2x-{{y}^{3}} \right)}^{7}}={{\sum\limits_{k=0}^{7}{\dfrac{7!}{\left( 7-k \right)!k!}.\left( 2x \right)}}^{7-k}}.{{\left( -{{y}^{3}} \right)}^{k}}$$ if we write formula as $${{\left( 2x \right)}^{k-7}}$$ instead of $${{\left( 2x \right)}^{7-k}}$$ our whole simplification makes our solution a wrong one.