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andrew-mc [135]
3 years ago
12

Hello, could anyone please help me with my homework?

Mathematics
1 answer:
vredina [299]3 years ago
4 0

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!

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