Question

In the expansion of ${\left(x^3-\frac{1}{x^2}\right)}^2$, where n is a positive integer, the sum of the coefficients of $x^6$ is $1.$

Answer 1

$(x^3-\frac{1}{x^2}{)}^2$

$=(x^3{)}^2-2(x^3)(\frac{1}{x^2})+(\frac{1}{x^2}{)}^2$ ................. Using, $(a-b{)}^2=a^2-2ab+b^2$

$=x^6-2x+\frac{1}{x^4}$

Sum of coefficients of $x^6$ is $1$.