Expand (1+X+2X^2)^6 as far as the term x^3?

= (1 + x + 2x²)⁶

= [(1 + x + 2x²)²]³

= [({1 + x} + 2x²)²]³

= [{1 + x}² + 4x².{1 + x} + 4x⁴]³ → no more x³, so you simplify

= [{1 + x}² + 4x².{1 + x}]³

= [1 + 2x + x² + 4x² + 4x³]³

= [1 + 2x + 5x² + 4x³]³

= [1 + 2x + 5x² + 4x³]².[1 + 2x + 5x² + 4x³]

= [(1 + 2x) + (5x² + 4x³)]².[1 + 2x + 5x² + 4x³]

= [(1 + 2x)² + 2.(1 + 2x).(5x² + 4x³) + (5x² + 4x³)²].[1 + 2x + 5x² + 4x³]

= [1 + 4x + 4x² + 10x² + 8x³ + 20x³ + 16x⁴ + 25x⁴ + 40x⁵ + 16x⁶].[1 + 2x + 5x² + 4x³]

= [1 + 4x + 14x² + 28x³ + 41x⁴ + 40x⁵ + 16x⁶].[1 + 2x + 5x² + 4x³] → no more x³, so you simplify

= [1 + 4x + 14x² + 28x³].[1 + 2x + 5x² + 4x³]

= 1.[1 + 2x + 5x² + 4x³] + 4x.[1 + 2x + 5x² + 4x³] + 14x².[1 + 2x + 5x² + 4x³] + 28x³.[1 + 2x + 5x² + 4x³] → no more x³, so you simplify

= 1 + 2x + 5x² + 4x³ + 4x.[1 + 2x + 5x²] + 14x².[1 + 2x] + 28x³.[1]

= 1 + 2x + 5x² + 4x³ + 4x + 8x² + 20x³ + 14x² + 28x³ + 28x³

= 1 + 6x + 27x² + 80x³

Using binomial expansion, (1 + x + 2x²)⁶
= [(1 + x) + 2x²]⁶
= (1 + x)⁶ + ₆C₁*(1 + x)⁵*(2x²) + ……
= (1 + ₆C₁*x + ₆C₂*x² + ₆C₃*x³ + ……) + 6*(1 + ₅C₁*x + ……)*(2x²) + ……
= (1 + 6x + 15x² + 20x³ + …… ) + 12x²*(1 + 5x + ……) + ……
= (1 + 6x + 15x² + 20x³ + …… ) + (12x² + 60x³ + ……) + ……
= 1 + 6x + 27x² + 80x³ + ……

(In expansion, the terms resulted in higher than x³ are shown as "……".)

Conveniently groping the given one is
= [1 + x(1 + 2x)]⁶

Expanding the above, using Binomial expansion,
= 1 + C1*{x(1 + 2x} + C2*{x²(1 + 2x)²} + C3*{x³(1 + 2x)³} + ......
[All other terms above this are not required, since we require only terms upto the power of x³]

In the above, C1 = 6; C2 = (6*5)/(1*2) = 15; C3 = (6*5*4)/(1*2*3) = 20

So of the above, the expansion
= 1 + 6x(1 + 2x) + 15x²(1 + 4x + 4x²) + 20x³(1 + 6x + 12x² + 8x³)
= 1 + 6x + 12x² + 15x² + 60x³ + 20x³ [Only the terms containing upto power x³ is considered]
= 1 + 6x + 27x² + 80x³ + .......

80 X^3 + 27 X^2 + 6 X + 1

64 X^12 + 192 X^11 + 432 X^10 + 640 X^9 + 780 X^8 + 732 X^7 + 581 X^6 + 366 X^5 + 195 X^4 + 80 X^3