Math, asked by nehab9457, 11 months ago

Expand (1+x+x²)³using bionomial theorem

Answers

Answered by IamIronMan0
1

Answer:

Let

x +  {x}^{2}  =x(1 + x) =  y

Now

(1 + y) {}^{3}  =  \binom{3}{0}  {y}^{0}  + \binom{3}{1}  {y}^{1}  + \binom{3}{2}  {y}^{2}  + \binom{3}{3}  {y}^{3}  \\  \:  \:  \:  \:   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   = 1 + 3y + 3 {y}^{2}  +  {y}^{3}

Substitute value of y

 = 1 + 3x(1 + x) + 3 {x}^{2} (1 + x) {}^{2}  + x^3 {(1 + x)}^{3}

Again use binomial

 = 1 + 3x(1 + x) + 3 {x}^{2} ( {x }^{2}  + x + 1) \\  \:  \:  \:  \:  \:  \:  \:  \:  +  {x}^{3} \bigg \{ \binom{3}{0}  { x}^{0}  + \binom{3}{1}  {x}^{1}  + \binom{3}{2}  {x}^{2}  + \binom{3}{3}  {x}^{3} \bigg \}\\   \\  = 1 + 3x + 3 {x}^{2}  + 3{x}^{4}  + 3 {x}^{3}  + 3 {x}^{2}  \\  \:  \:  \:   \:  \:  \:  \:  \: +  {x}^{3}  +  3 {x}^{4} + 3 {x}^{5}  +  {x}^{6}  \\  =  {x}^{6}  + 3 {x}^{5}  + 6 {x}^{4}  + 4 {x}^{3}  + 6 {x}^{2}  + 3x + 1

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