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

1 Answer
May 17, 2017

Answer:

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

Explanation:

Binomial expansion of #(a-b)^n# is given by

#C_0^na^nb^0+C_1^na^(n-1)b^1+C_2^na^(n-2)b^2+C_3^na^(n-3)b^3+......+C_r^na^(n-r)b^r+..........+C_n^na^0b^n#, where #C_r^n=(n!)/((n-r)!r!)#.

Hence #(x^2+y^2)^6#

= #C_0^6(x^2)^6(y^2)^0+C_1^6(x^2)^(6-1)(y^2)^1+C_2^6(x^2)^(6-2)(y^2)^2+C_3^6(x^2)^(6-3)(y^2)^3+C_4^6(x^2)^(6-4)(y^2)+C_5^6(x^2)^(6-5)(y^2)^5+C_6^n(x^2)^0(y^2)^6#

= #x^12+6/1xx x^10y^2+(6*5)/(1*2)x^8y^4+(6*5*4)/(1*2*3)x^6y^6+(6*5*4*3)/(1*2*3*4)x^4y^8+(6*5*4*3*2)/(1*2*3*4*5)x^2y^10+y^12#

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