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

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

Explanation

Solution

We first define the general form of binomial expansion for the indices value of n where (a−b)n=an−nC1an−1b1+nC2an−2b2−.....+(−1)rnCran−rbr+.....+(−1)nbn{{\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=1a=3u,b=1 and n=5n=5. Then we use the formula of combinational nCr=n!r!×(n−r)!{}^{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
(a−b)n=an−nC1an−1b1+nC2an−2b2−.....+(−1)rnCran−rbr+.....+(−1)nbn{{\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 tr+1{{t}_{r+1}}, the (r+1)th{{\left( r+1 \right)}^{th}} term of the series where tr+1=(−1)rnCran−rbr{{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=5n=5.
We also replace the values for a=3u,b=1a=3u,b=1.
We put the values in the main equation of expansion and get
(3u−1)5=(3u)5−5C1(3u)5−1+5C2(3u)5−2−5C3(3u)5−3+5C4(3u)5−4−1{{\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 nCr=n!r!×(n−r)!{}^{n}{{C}_{r}}=\dfrac{n!}{r!\times \left( n-r \right)!}.
Putting the respective values, we get 5C2=5C3=5!2!×3!=10;5C1=5C4=5{}^{5}{{C}_{2}}={}^{5}{{C}_{3}}=\dfrac{5!}{2!\times 3!}=10;{}^{5}{{C}_{1}}={}^{5}{{C}_{4}}=5.
Therefore, the expansion becomes
(3u−1)5=243u5−405u4+270u3−90u2+15u−1{{\left( 3u-1 \right)}^{5}}=243{{u}^{5}}-405{{u}^{4}}+270{{u}^{3}}-90{{u}^{2}}+15u-1.
The expansion form of (3u−1)5{{\left( 3u-1 \right)}^{5}} is 243u5−405u4+270u3−90u2+15u−1243{{u}^{5}}-405{{u}^{4}}+270{{u}^{3}}-90{{u}^{2}}+15u-1.

Note: We can also use the concept of (3u−1)5=(3u−1)3(3u−1)2{{\left( 3u-1 \right)}^{5}}={{\left( 3u-1 \right)}^{3}}{{\left( 3u-1 \right)}^{2}}. Then we use the cubic and quadratic formulas of (a−b)3=a3−3a2b+3ab2−b3{{\left( a-b \right)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}} and a2−2ab+b2=(a−b)2{{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}} respectively. We multiply the values of the expansion to get the same solution.