Question

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

Answer 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

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