Expand $(1 + x + x^{2})^{3}$ using binomial expansion.

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Solution

Put 1 + x = y. then,
$(1 + x + x^{2})^{3} = (y + x^{2})^{3}$
$=^{3} C_{0} y^{3} +^{3} c_{1} y^{2} x^{2} +_{2}^{C} y (x^{2})^{2} +^{3} C_{3} (x^{2})^{3} = y^{3} + 3 y^{2} x^{2} + 3 y x^{4} + x^{6} = (1 + x)^{3} + 3 x^{2} (1 + x)^{2} + 3 x^{4} (1 + x) + x^{6} = (1 +^{3} C_{1} x +^{3} C_{2} x^{2} +^{3} C_{3} x^{3}) + 3 x^{2} (1 + 2 x + x^{2}) + 3 x^{4} (1 + x) + x^{6} = (1 + 3 x + 3 x^{2} + x^{3}) + (3 x^{2} + 6 x^{3} + 3 x^{4}) + (3 x^{4} + 3 x^{5}) + x^{6} = x^{6} + 3 x^{5} + 6 x^{4} + 7 x^{3} + 6 x^{2} + 3 x + 1. Hence (1 + x + x^{2}) 3^{=} x^{6} + 3 x^{5} + 6 x^{4} + 7 x^{3} + 6 x^{2} + 3 x + 1.$

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