Question

Hello, could anyone please help me with my homework?

Answer 1

Answer:

$(x + 6 {)}^{6} = {x}^{6 } + 36{x}^{5} + 540{x}^{4} + 4320 {x}^{3} + 19440{x}^{2}+ 46656{x}^{} + 46656$

Step-by-step explanation:

we want to solve the following binomial:

$(x + 6 {)}^{6}$

There is a handy way to expand powers of binomials which is known as binomial theorem . and it describes the algebraic expansion of powers of a binomial. binomial theorem is given by

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

comparing (x+6)⁶ to (a+b)ⁿ , we get

• $a \implies \: x$
• $b\implies \: 6$
• $n\implies \: 6$

now substitute them on the formula which yields:

$\displaystyle (x + 6 {)}^{6} = \sum _{k = 0 }^{n} \binom{6}{k} {x}^{6 - k} \cdot {6}^{k}$

converting the summation notation into sum yields:

$\displaystyle (x + 6 {)}^{6} = \binom{6}{0} {x}^{6 - 0} \cdot {6}^{0} + \binom{6}{1} {x}^{6 - 1} \cdot {6}^{1} + \binom{6}{2} {x}^{6 - 2} \cdot {6}^{2} + \binom{6}{3} {x}^{6 - 3} \cdot {6}^{3} + \binom{6}{4} {x}^{6 - 4} \cdot {6}^{4} + \binom{6}{5} {x}^{6 - 5} \cdot {6}^{5} + \binom{6}{6} {x}^{6 - 6} \cdot {6}^{6} \\ \implies(x + 6 {)}^{6} = \binom{6}{0} {x}^{6 } \cdot 1 + \binom{6}{1} {x}^{5} \cdot {6} + \binom{6}{2} {x}^{4} \cdot 36+ \binom{6}{3} {x}^{3} \cdot 216 + \binom{6}{4} {x}^{2} \cdot 1296+ \binom{6}{5} {x}^{} \cdot 7776+ \binom{6}{6} {x}^{0} \cdot 46656 \\ \implies(x + 6 {)}^{6} = 1 \cdot {x}^{6 } \cdot 1 + 6 \cdot{x}^{5} \cdot {6} + 15 \cdot {x}^{4} \cdot 36+ 20 \cdot {x}^{3} \cdot 216 + 15 \cdot{x}^{2} \cdot 1296+ 6 \cdot {x}^{} \cdot 7776+ 1 \cdot {x}^{0} \cdot 46656 \\ \implies \boxed{ (x + 6 {)}^{6} = {x}^{6 } + 36{x}^{5} + 540{x}^{4} + 4320 {x}^{3} + 19440{x}^{2}+ 46656 {x}^{} + 46656 }$

and we're done!