The sum of the coefficients of first three term in the expansion of `(x-3/(x^2))^m\, x!=0.` m being a natural number, is 559. Find the term of the expansion containing `x^3`.

Explanation

Sum of coefficients of the first three terms of `(x-frac{3}{x^{2}})^{m}`

Now we know (- in the expansion `(a+b)^{n}`,

the general term is `T_{r+1}=\ ^{n} C_{r} a^{n-T} b^{r}`

In expansion `(x-frac{3}{x^{2}})^{m}`, general term is `T_{r+1}=m C_{T} x^{m-r}(frac{-3}{x^{2}})^{r}`

Now, first 3 terms are `T_{1}, T_{2}` and `T_{3}`

`T_{1}=\ ^{m} C_{0} x^{m}(frac{-3}{x^{2}})^{0}=x^{m} `

`T_{2}=\ ^{m} C_{1} x^{m-1}(frac{-3}{x^{2}})^{1}=-3^{m} C_{1} x^{m-3}=-3^{3} C_{1} `

`T_{3}=\ ^{m} C_{2}(x^{m-2})^{-3}(frac{-3}{x^{2}})^{2}=\ ^{m} C_{2} x^{m-2}(frac{(-3)^{2}}{x^{4}})=9^{m} C_{2} x^{m-6} `

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