This will be the #3#rd term of the expansion. This is because we always start #x# at the highest power and #y# at the lowest. Also, there will be #8 + 1 = 9# terms in the expansion, so the term of the form #ax^6y^2# will be the #9 - 6 = 3#rd term.
The formula #t_(k + 1) = color(white)(two)_nC_k(x)^(n - k)y^k# will let you find the term that you're looking for of #(x + y)^n#.
#3 = k+ 1 -> k = 2#
Now, substitute your known values into the formula.
#t_3 = color(white)(two)_8C_2 (2x)^(8 - 2)(-3y)^2#
We now notice that #a = 64 * 9 *color(white)(two)_8C_2#. We can evaluate this using the combination formula, which states that #color(white)(two)_nC_r = (n!)/(r!(n - r)!)#, where #n! = n(n - 1)(n - 2)(...)(1)#.
#a = (8!)/(2!(8 - 2)!) * 64 * 9#
#a = (8!)/(6!2!) * 64 * 9#
#a = (8 * 7 * 6!)/(6! * 2!) * 64 * 9#
#a = 56/2 * 64 * 9#
#a = 28* 64 * 9#
#a = 16,128#
Hopefully this helps!
