Binomial Theorem

The coefficient of $x^2$ corresponds to $k=2$: $\left(\genfrac{}{}{0px}{}{4}{2}\right)=6$.

$(x-1{)}^3=x^3-3x^2+3x-1$.

The binomial formula gives $(a+b{)}^2=a^2+2ab+b^2$.

The general term in the expansion is $$T_{k+1}=\left(\genfrac{}{}{0px}{}{5}{k}\right)(2x{)}^{5-k}(3{)}^k$$ The power of $x$ is $5-k$, so set $$5-k=3⟹k=2$$ Substituting $k=2$: $$\left(\genfrac{}{}{0px}{}{5}{2}\right)(2{)}^3(3{)}^2=10\cdot 8\cdot 9=720$$

${\sum }_{k=0}^6(-1{)}^k\left(\genfrac{}{}{0px}{}{6}{k}\right)=(1-1{)}^6=0$.

$5-k=3\Rightarrow k=2$. Coefficient: $\left(\genfrac{}{}{0px}{}{5}{2}\right)\cdot 2^2=10\cdot 4=40$.

With $a=2x$ and $b=3$: $T_{k+1}=\left(\genfrac{}{}{0px}{}{n}{k}\right)(2x{)}^{n-k}{3}^k$.

The sum of coefficients of a polynomial $P(x)$ is $P(1)$. So ${\sum }_{k=0}^{10}\left(\genfrac{}{}{0px}{}{10}{k}\right)=(1+1{)}^{10}=2^{10}=1024$.

The general term in the expansion is $$\left(\genfrac{}{}{0px}{}{5}{r}\right)(2{)}^{5-r}(3x{)}^r\mathrm{.}$$

The term in $x^3$ comes from $r=3$: $$\left(\genfrac{}{}{0px}{}{5}{3}\right)(2{)}^{5-3}(3{)}^3{x}^3\mathrm{.}$$

Evaluate each factor: $$\left(\genfrac{}{}{0px}{}{5}{3}\right)=10,2^2=4,3^3=27.$$

Multiply: $$10\times 4\times 27=1080.$$

So the coefficient of $x^3$ is $1080$.

The general term is $$T_k=\left(\genfrac{}{}{0px}{}{5}{k}\right)(2x{)}^{5-k}(3{)}^k\mathrm{.}$$ For the $x^3$ term, set the exponent of $x$ equal to $3$: $$5-k=3⟹k=2.$$ Substitute $k=2$: $$\left(\genfrac{}{}{0px}{}{5}{2}\right)(2x{)}^3(3{)}^2=10\cdot 8x^3\cdot 9.$$ So the coefficient is $$10\cdot 8\cdot 9=720.$$