Question

Expand the function (1 + x + x²)³ using the Trinomial theorem​

Answer 1

Answer:

its hard

Step-by-step explanation:

Answer 2

Step-by-step explanation:

(a+b)² = a²+b²+2ab

(a-b)³ = a³-b³-3a²b+3ab²

(x²+x+1)³

finding the roots of x²+x+1

where the roots are w,w²(w=omega)

where w =

$( - 1 + \sqrt{3} ) \div 2 = w$

$( - 1 - \sqrt{3} ) \div 2 = {w}^{2}$

so x²+x+1 can be written as (x-w)(x-w²)

then (1 + x + x²)³ = ((x-w)(x-w²))³

= (x-w)³(x-w²)³

(x-w)³= x³-1-3x²w+3xw²

(x-w²)³ = x³-1-3x²w+3xw

=(x³-1-3x²w+3xw²)(x³-1-3x²w+3xw)