Answer
$35a^4b^3$
Work Step by Step
$(a+b)^n = a^n + n/1!* a^{n-1}*b^1 + n(n-1)/2!* a^{n-2}*b^2 + n(n-1)(n-2)/3!* a^{n-3}*b^3 + . . . + b^n$
$a=a$
$b=b$
$n=7$
$(a+b)^n = a^n + 7/1!* a^{7-1}*b^1 + 7(7-1)/2!* a^{7-2}*b^2 + 7(7-1)(7-2)/3!* a^{7-3}*b^3 + . . . + b^n$
$7(7-1)(7-2)/3!* a^{7-3}*b^3$ is the 4th term
$7(7-1)(7-2)/3!* a^{7-3}*b^3$
$7(7-1)(7-2)/6*a^4b^3$
$7*6*5/6*a^4b^3$
$35a^4b^3$
