How Do you expand (4x+3y+3z)^6?
To expand (4x+3y+3z)^6, you can use the binomial theorem which states that for any binomial expression raised to a positive integer power, the expansion can be determined using the coefficients of the terms.
The expansion can be written as:
(4x+3y+3z)^6 = C(6,0)(4x)^6(3y)^0(3z)^0 + C(6,1)(4x)^5(3y)^1(3z)^0 + C(6,2)(4x)^4(3y)^2(3z)^0 + C(6,3)(4x)^3(3y)^3(3z)^0 + C(6,4)(4x)^2(3y)^4(3z)^0 + C(6,5)(4x)^1(3y)^5(3z)^0 + C(6,6)(4x)^0(3y)^6(3z)^0
where C(n,r) denotes the binomial coefficient, which is calculated by the formula C(n,r) = n! / (r!(n-r)!), and (n!) denotes the factorial of a number n.
Calculating each term step by step:
Term 1: C(6,0)(4x)^6(3y)^0(3z)^0 = 1*(4x)^6*(3y)^0*(3z)^0 = 4096x^6
Term 2: C(6,1)(4x)^5(3y)^1(3z)^0 = 6*(4x)^5*(3y)^1*(3z)^0 = 1152x^5y
Term 3: C(6,2)(4x)^4(3y)^2(3z)^0 = 15*(4x)^4*(3y)^2*(3z)^0 = 2160x^4y^2
Term 4: C(6,3)(4x)^3(3y)^3(3z)^0 = 20*(4x)^3*(3y)^3*(3z)^0 = 2160x^3y^3
Term 5: C(6,4)(4x)^2(3y)^4(3z)^0 = 15*(4x)^2*(3y)^4*(3z)^0 = 1620x^2y^4
Term 6: C(6,5)(4x)^1(3y)^5(3z)^0 = 6*(4x)^1*(3y)^5*(3z)^0 = 729xy^5
Term 7: C(6,6)(4x)^0(3y)^6(3z)^0 = 1*(4x)^0*(3y)^6*(3z)^0 = 729y^6
Therefore, the expanded form of (4x+3y+3z)^6 is:
4096x^6 + 1152x^5y + 2160x^4y^2 + 2160x^3y^3 + 1620x^2y^4 + 729xy^5 + 729y^6