Question

How do you expand ${{\left( 3u-1 \right)}^{5}}$?

Answer 1

Hint: We first define the general form of binomial expansion for the indices value of n where ${{\left( a-b \right)}^{n}}={{a}^{n}}-{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}-.....+{{\left( -1 \right)}^{r}}{}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}+.....+{{\left( -1 \right)}^{n}}{{b}^{n}}$. We replace the values with $a=3u,b=1$ and $n=5$. Then we use the formula of combinational ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\times \left( n-r \right)!}$ to find the coefficients. We put the values and get the final solution for the expansion.

Complete step-by-step answer:
We use the formula for binomial expansion where we have
${{\left( a-b \right)}^{n}}={{a}^{n}}-{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}-.....+{{\left( -1 \right)}^{r}}{}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}+.....+{{\left( -1 \right)}^{n}}{{b}^{n}}$.
The general term of the expansion is ${{t}_{r+1}}$, the ${{\left( r+1 \right)}^{th}}$ term of the series where ${{t}_{r+1}}={{\left( -1 \right)}^{r}}{}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}$.
We use the binomial expansion for the value of $n=5$.
We also replace the values for $a=3u,b=1$.
We put the values in the main equation of expansion and get
\[{{\left( 3u-1 \right)}^{5}}={{\left( 3u \right)}^{5}}-{}^{5}{{C}_{1}}{{\left( 3u \right)}^{5-1}}+{}^{5}{{C}_{2}}{{\left( 3u \right)}^{5-2}}-{}^{5}{{C}_{3}}{{\left( 3u \right)}^{5-3}}+{}^{5}{{C}_{4}}{{\left( 3u \right)}^{5-4}}-1\].
Now we know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\times \left( n-r \right)!}$.
Putting the respective values, we get \[{}^{5}{{C}_{2}}={}^{5}{{C}_{3}}=\dfrac{5!}{2!\times 3!}=10;{}^{5}{{C}_{1}}={}^{5}{{C}_{4}}=5\].
Therefore, the expansion becomes
\[{{\left( 3u-1 \right)}^{5}}=243{{u}^{5}}-405{{u}^{4}}+270{{u}^{3}}-90{{u}^{2}}+15u-1\].
The expansion form of ${{\left( 3u-1 \right)}^{5}}$ is \[243{{u}^{5}}-405{{u}^{4}}+270{{u}^{3}}-90{{u}^{2}}+15u-1\].

Note: We can also use the concept of ${{\left( 3u-1 \right)}^{5}}={{\left( 3u-1 \right)}^{3}}{{\left( 3u-1 \right)}^{2}}$. Then we use the cubic and quadratic formulas of ${{\left( a-b \right)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}}$ and ${{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}$ respectively. We multiply the values of the expansion to get the same solution.