Binomial expansion, where is the mistake?

Question

I attacked the Question and the marking scheme Question 3.

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Question Find the term independent of $x$ in the series expansion of $$\left(4x^3+\frac1{2x}\right)^8$$

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First I expanded inside the bracked, then took the 2x out and got: $$\dfrac{1}{256x^8}(4x^4+1)^8$$ Now from here we can see that we get $x^8$ when $(4x^4)$ will be on the power of 2, which is at $8C6$ Now that will be: $$8C6(\dfrac{1}{256x^8})(4x^4)^2$$ which is $$28*16x^8(\dfrac{1}{256x^8})$$ which will be $$\dfrac{7}{4}$$ but the marking scheme shows just 7. Where did I make a mistake?

Answer 1

When you took $2x$ out, it would be $$\frac{1}{256x^8}(8x^4+1)^8$$ instead of $$\dfrac{1}{256x^8}(4x^4+1)^8$$

Now, your answer will be multiplied by a factor of$2^2$ which will finally give the answer to be $7$.

Answer 2

The binomial expansion will have a constant term when the powers of $x$ $$ \left( 4x^3\right)^{p} \left( \frac{1}{2x} \right)^{q} $$ total to $0$. That is $$ x^{3p-q} = x^{0}. $$ Therefore $$ q = 3p. \tag{1} $$ Each term in the binomial expansion has order 8. Using Pascal's triangle $$ \left( 4x^{3} \right)^{8} + 8 \left( 4x^{3} \right)^{7}\left( \frac{1}{2x} \right)^{1} + \color{blue}{28 \left( 4x^{3} \right)^{6} \left( \frac{1}{2x} \right)^{2}} + \dots $$ In other words, $$ p + q = 8 \tag{2} $$ Which values of $p$ and $q$ satisfy (1) and (2)? $$ p =2, \quad q = 6. $$ This is $\color{blue}{blue}$ term in the expansion: $$ \boxed{ \color{blue}{28 \left( 4x^{3} \right)^{6} \left( \frac{1}{2x} \right)^{2}} = 28 \frac{16}{64} = 7 } $$

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Pascal's triangle