How do you use the binomial theorem to expand and simplify the expression #(x^2+y^2)^6#?

1 Answer
Jul 7, 2017

Answer:

# (x^2+y^2)^6 = x^12 + 6 x^10 y^2 + 15 x^8 y^4 + 20 x^6 y^6 + 15 x^4 y^8 + 6 x^2 y^10 + y^12 #

Explanation:

Recall the Binomial Theorem

# (a+b)^n = sum_(r=0)^n \ ( (n), (r) ) \ a^(n-r) \ b^r #

Where:

# ( (n), (r) ) =""^nC_r = (n!)/((n-r)!r! #

is the combinatorial, which are also the numbers from the #nth# row of Pascal's Triangle

Thus we have:

# (x^2+y^2)^6 = sum_(r=0)^6 \ ( (6), (r) ) \ (x^2)^(6-r) \ (y^2)^r #

# " " = ( (6), (0) ) (x^2)^(6) + ( (6), (1) ) (x^2)^(5) (y^2) + ( (6), (2) ) (x^2)^(4) (y^2)^(2) + ( (6), (3) ) (x^2)^(3) (y^2)^(3) + ( (6), (4) ) (x^2)^(2) (y^2)^(4) + ( (6), (5)) (x^2) (y^2)^(5) + ( (6), (6)) (y^2)^(6) #

# " " = (1) (x^12) + (6) (x^10) (y^2) + (15) (x^8) (y^4) + (20) (x^6) (y^6) + (15) (x^4) (y^8) + (6) (x^2) (y^10) + (1) (y^12) #

# " " = x^12 + 6 x^10 y^2 + 15 x^8 y^4 + 20 x^6 y^6 + 15 x^4 y^8 + 6 x^2 y^10 + y^12 #