Binomial Problem : Find the coefficient of this equation!

Question

How to find $x^{13}$ from this equation:

$\left ( x^{3}+1 \right )^{2}\left ( x^{2}-\frac{2}{x} \right )^{8}$

I'm very confused with these equation and don't know how to solve using Binomial Newton Theorem....

Answer 1

Hint: Your expression can be rewritten as $$\left ( x^{6}+2x^3+1 \right )\left ( x^{2}-\frac{2}{x} \right )^{8}$$ Now you can manually find and add required coefficients $$x^{13}=x^{6}\times x^{7}=x^{3}\times x^{10}=x^{0}\times x^{13}$$ Your answer will be then

Coeficient of $x^7$ in $\left ( x^{2}-\frac{2}{x} \right )^{8}$+$2\times$Coeficient of $x^{10}$ in $\left ( x^{2}-\frac{2}{x} \right )^{8}$+Coeficient of $x^{13}$ in $\left ( x^{2}-\frac{2}{x} \right )^{8}$ $$=-448+2\times112-16=-240$$

Answer 2

You could start with $\left ( x^{3}+1 \right )^{2}\left ( x^{2}-\frac{2}{x} \right )^{8} = \frac{1}{x^8} \left ( x^{3}+1 \right )^{2}\left ( x^{3}-2 \right )^{8}$ so your question is equivalent to finding the coefficient of $x^{21}$ in $\left ( x^{3}+1 \right )^{2}\left ( x^{3}-2 \right )^{8}$.

Letting $y=x^3$, this is the same as finding the coefficient of $y^7$ in $\left ( y+1 \right )^{2}\left ( y-2 \right )^{8}$.

This requires some calculating, but when expanding $\left( y-2 \right )^{8}$ you can ignore any powers smaller than $y^5$.

Answer 3

To supplement the excellent answers provided earlier by others, here's a table which might be helpful. The table shows that powers of $x$ from the expansion of the first and second terms are $6, 3, 0$, and $16, 13, 10,...,-8$ respectively. The power 13 is a resultant of the product of powers $(6,3,0)$ with $(7,10,13)$ respectively, and its coefficient is the sum of coefficients of the corresponding products, as shown in the highlighted area.