Answer
The 3rd term is equal to $189x^5$.
Work Step by Step
According to the binomial theorem, we can expand the algebraic expression in the form of:
$(a+b)^n=\binom{n}{0}b^0a^n+\binom{n}{1} b^1a^{n-1}++\binom{n}{n-i} b^{n-i}a^{i}++\binom{n}{n-1}b^{n-1}a^1+\binom{n}{n}b^nx^0$
Now, we will expand the given expression by replacing $(a+b)^n$ with $(x-3)^7$.
$(x-3)^7=\dbinom{7}{0}(-3)^0x^7+\dbinom{7}{1}(-3)^1x^6+\dbinom{7}{2}(-3)^2x^5+\dbinom{7}{3}(-3)^3x^4+\dbinom{7}{4}(-3)^4x^3+\dbinom{7}{5}(-3)^5x^2+\dbinom{7}{6}(-3)^6x^1+\dbinom{7}{7}(-3)^7x^0$
Thus, the 3rd term is equal to $\dbinom{7}{2}(-3)^2x^5=21\times 9 x^5=189x^5$.
