Let us study the concept by finding the constant term in the expansion of (x^{3} – 1/x^{2})^{15}

In (a + b)^{n} the (r + 1)^{th} term T_{r+1} is given by –

T_{r+1} = ^{n}C_{r} .a^{n-r}. b^{r} for r = 0, 1, 2, 3, 4, 5 … n

Let constant terms occurs at (r+1)^{th} term for ( x^{3} – 1/x^{2} )^{15}

So,

T_{r+1} = ^{n}C_{r} a^{n-r}.b^{r} = ^{15}C_{r} (x^{3})^{15-r}.(x^{-2})^{r}

T_{r+1} = ^{15}C_{r} (x)^{45-5r}

For constant term, x should be eliminated, so –

45 -5r = 0

r = 9

So the constant term is T_{9+1} , so T_{10}

T_{10} = ^{15}C_{9} (x)^{45-45}

T_{10} = ^{15}C_{9}