the constant term in the expansion of ((3/x^2)-4x^3)^10

$$\left(\dfrac{3}{x^2}-4x^3\right)^{10}$$

    $$\mbox{The (r+1)th term will be }\quad 0\le{r}\le10\quad r\in \mathBb{Z}$$       

$$10Cr\times \left(\frac{3}{x^2}\right)^r\times (-4x^3)^{10-r}\\ =10Cr \times 3^r\times(-4)^{10-r}\times \dfrac{1}{x^{2r}}\times (x^3)^{10-r}\\ =10Cr \times 3^r\times(-4)^{10-r}\times x^{-2r}\times x^{30-3r}\\ =10Cr \times 3^r\times(-4)^{10-r}\times x^{-2r+30-3r}\\ =10Cr \times 3^r\times(-4)^{10-r}\times x^{30-5r}$$  

You want the constant term so 30-5r=0,   r=6  

so the term we want is 

$$10C6\times 3^6\times (-4)^4 \times x^0$$

etc (correct assuming I didn't make any stupid errors)

Math(Input=Result) Error!

$${binom}{\left({\left({\frac{{\mathtt{3}}}{{{\mathtt{x}}}^{{\mathtt{2}}}}}{\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{3}}}\right)}^{{\mathtt{10}}}\right)} = {binom}{\left(\tiny\text{Error: loop}\right)}$$

$$\left(\dfrac{3}{x^2}-4x^3\right)^{10}\\$$

I think the answer is:

$$10Cr\times \left(\dfrac{3}{x^2}\right)^{\textcolor[rgb]{1,0,0}{10-r}}\times (-4x^3)^{\textcolor[rgb]{1,0,0}{r}}\\ =10Cr\times 3^{10-r} \times (-4)^r \times x^{-2(10-r)} \times x^{3r}\\ =10Cr\times 3^{10-r} \times (-4)^r \times x^{-2(10-r)+3r}\\ =10Cr\times 3^{10-r} \times (-4)^r \times x^{-20+2r+3r}\\ =10Cr\times 3^{10-r} \times (-4)^r \times x^{-20+5r}$$

so -20+5r=0  -> r = 4

the term we want is

$$10C\textcolor[rgb]{1,0,0}{4}\times 3^6\times (-4)^4 \times x^0$$

Using sigma notation and factorials for the combinatorial numbers, here is the binomial theorem:

Hi Heureka,

Mine is not wrong.

nCr is always identical to in value to nC(n-r)

10C4=10C6   

your answer and my answer are identical.

Hi Melody,

your answer is not wrong!

But r ist also a position in the Pascal triangle too.

$$\\\mbox{r = 0: } (\frac{3}{x^2})^{10}=\dfrac{59049}{x^{20}}\\ \mbox{r = 1: } \dfrac{ -787320*x^{5} }{x^{20}}\\ \mbox{r = 2: } \dfrac{ 4723920*x^{10} }{x^{20}}\\ \mbox{r = 3: } \dfrac{ -16796160*x^{15} }{x^{20}}\\ \textcolor[rgb]{1,0,0}{\mbox{r = 4: } \dfrac{ 39191040*x^{20} }{x^{20}} = 39191040} \\ \mbox{r = 5: } \dfrac{ -62705664*x^{25} }{x^{20}}\\ \mbox{r = 6: } \dfrac{ 69672960*x^{30} }{x^{20}}\\ \mbox{r = 7: } \dfrac{ -53084160*x^{35} }{x^{20}}\\ \mbox{r = 8: } \dfrac{ 26542080*x^{40} }{x^{20}}\\ \mbox{r = 9: } \dfrac{ -7864320*x^{45} }{x^{20}}\\ \mbox{r = 10: } (-4x^3)^{10}=\dfrac{ 1048576*x^{50} }{x^{20}}$$

Many Greetings

Heureka

They are identical Heureka.

$$\left(\dfrac{3}{x^2}-4x^3\right)^{10}=\left(-4x^3+\dfrac{3}{x^2}\right)^{10}$$

You can start the expansion from either end.  It makes no difference.