A father and his three children decide to hold a vote to select an after dinner activity. They will either see a play or a movie. If the majority of the family agrees on a preferred activity, then they will do that activity. If two family members vote to see a play and the other two family members vote to watch a movie, then (since the father will pay for the activity), the father's vote will be used to break the tie. (a) How many winning coalitions are there in this situation?
(b) Find one of the children's Banzhaf power index.

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75%

step-by-step explanation:

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)