Answers: 1
is 
.
.
and 
as the numbers and the binomial expression becomes 
. If the power is assumed to be 
, then the binomial expression is 
.
, the expression is 
where, 
. ![\begin{aligned}(d-4b)^{3}&=\left[\dbinom{3}{0}(d)^{3}(-4b)^{0}+\dbinom{3}{1}(d)^{2}(-4b)^{1}+\dbinom{3}{2}(d)^{1}(-4b)^{2}+\dbinom{3}{3}(d)^{0}(-4b)^{3}+\right]\\&=\left[\dfrac{3!}{3!\cdot 0!}d^{3}(1)+\dfrac{3!}{1!\cdot 2!}(-d^{2}4b)+\dfrac{3!}{2!\cdot 1!}d(16b^{2})+\dfrac{3!}{3!\cdot 0!}d(1)(-64b^{3})\right]\end{aligned}](/tpl/images/0396/2397/7f6cc.png)
![\begin{aligned}{\left({d - 4b}\right)^3}&=\left[{{d^3} + 3\left({ - {d^2}4b} \right) + 3d\left( {16{b^2}}\right)+\left({ - 64{b^3}}\right)} \right]\\&={d^3} - 12b{d^2} + 48{b^2}d - 64{b^3}\\\end{aligned}](/tpl/images/0396/2397/334c8.png)
is 
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Use the binomial theorem to expand (d-4b)^3...
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