Expand (1+x+x^2)^3
( 1 + x + x ^ 2 ) ^ 3 = [ ( 1 + x ) + x ^ 2 ] ^ 3
Now use formulas:
( a + b ) ^ 2 = a ^ 2 + 2 a b + b ^ 2
and
( a + b ) ^ 3 = a ^ 3 + 3 a ^ 2 * b + 3 a * b ^ 2 + b ^ 3
[ ( 1 + x ) + x ^ 2 ] ^ 3 =
( 1 + x ) ^ 3 + 3 ( 1 + x ) ^ 2 * x ^ 2 + 3 ( 1 + x ) * ( x ^ 2 ) ^ 2 + ( x ^ 2 ) ^ 3 =
( 1 + x ) ^ 3 + 3 ( 1 + x ) ^ 2 * x ^ 2 + 3 ( 1 + x ) * x ^ 4 + x ^ 6 =
( 1 ^ 3 + 3 * 1 ^ 2 * x + 3 * 1 * x ^ 2 + x ^ 3 ) + 3 * ( 1 ^ 2 + 2 * 1 * x + x ^ 2 ) * x ^ 2 + ( 3 + 3 x ) * x ^ 4 + x ^ 6 =
1 + 3 * 1 * x + 3 * x ^ 2 + x ^ 3 + 3 * ( 1 + 2 x + x ^ 2 ) * x ^ 2 + ( 3 + 3 x ) * x ^ 4 + x ^ 6 =
1 + 3 x + 3 x ^ 2 + x ^ 3 + ( 3 * 1 + 3 * 2 x + 3 * x ^ 2 ) * x ^ 2 + ( 3 * x ^ 4 + 3 x * x ^ 4 ) + x ^ 6 =
1 + 3 x + 3 x ^ 2 + x ^ 3 + ( 3 + 6 x + 3 x ^ 2 ) * x ^ 2 + 3 x ^ 4 + 3 x ^ 5 + x ^ 6 =
1 + 3 x + 3 x ^ 2 + x ^ 3 + ( x ^ 2 * 3 + x ^ 2 * 6 x + x ^ 2 * 3 x ^ 2 ) + 3 x ^ 4 + 3 x ^ 5 + 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 + 3 x ^ 4 + 3 x ^ 4 + 6 x ^ 3 + x ^ 3 + 3 x ^ 2 + 3 x ^ 2 + 3 x + 1 =
x ^ 6 + 3 x ^ 5 + 6 x ^ 4 + 7 x ^ 3 + 6 x ^ 2 + 3 x + 1
(a+b+c)^3 =
a^3 + 3a^2b + 3a^2c + 3ab^2 + 6abc + 3ac^2 + b^3 + 3b^2c + 3bc^2 + c^3
so, (x^2+x+1)^3 expands as
x^6 + 3x^5 + 3x^4 + 3x^4 + 6x^3 + 3x^2 + x^3 + 3x^2 + 3x + 1
which is what was above
To expand the expression (1+x+x^2)^3, we can use the binomial theorem or the method of distributing and simplifying. Let's use the method of distributing and simplifying:
To expand (1+x+x^2)^3, we need to multiply the expression by itself three times. Start by multiplying the first term, 1, by each term inside the parentheses:
(1+x+x^2) * (1+x+x^2) = 1 * (1+x+x^2) + x * (1+x+x^2) + x^2 * (1+x+x^2)
Next, distribute the terms and simplify:
= 1 + x + x^2 + x + x^2 + x^3 + x^2 + x^3 + x^4
Combine like terms:
= 1 + 2x + 3x^2 + 2x^3 + x^4
So, (1+x+x^2)^3 = 1 + 2x + 3x^2 + 2x^3 + x^4.