Answer (1 of 4) (xyz)^3 put xy = a (az)^3= a^3 z^3 3az ( az) = (xy)^3 z^3 3 a^2 z 3a z^2 = x^3y^3 z^3 3 x^2 y 3 x y^2 3(xy)^2 z 3(xy) z^2 =x^3 y^3 z^3 3 x^2y 3xy^2 3 ( x^2 y^2 2xy ) z 3x z^2 3yz^2 =x^3y^3z^3 3x^2 y3xy^2 3x^2 z 3y^2 z 6xyz 3xz^2 3= {2x ( y) z}²The Binomial Theorem is a formula that can be used to expand any binomial (xy)n =∑n k=0(n k)xn−kyk =xn(n 1)xn−1y(n 2)xn−2y2( n n−1)xyn−1yn ( x y) n = ∑ k = 0 n ( n k) x n − k y k = x n ( n 1) x n − 1 y ( n 2) x n − 2 y 2 ( n n − 1) x y n − 1 y n
General Expanded Form Of X Y Z K Mathematics Stack Exchange
