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ANSWER:
[tex](x-4y)^7=x^7-28x^6y+336x^5y^2\ldots[/tex]STEP-BY-STEP EXPLANATION:
We have the following expression:
[tex]\mleft(x-4y\mright)^7[/tex]In this case we can apply the binomial theorem, which is the following:
[tex](a+b)^n=\sum ^n_{i\mathop=0}(\frac{n!}{i!(n-i)!}a^{n-i}\cdot b^i[/tex]we replace and calculate for the first three terms:
[tex]\begin{gathered} 1st=\sum ^7_{i\mathop{=}0}(\frac{7!}{0!(7-0)!}x^{7-0}\cdot(-4y)^0=1\cdot x^7\cdot1=x^7 \\ 2nd=\sum ^7_{i\mathop{=}1}(\frac{7!}{1!(7-1)!}x^{7-1}\cdot(-4y)^1=7\cdot x^6\cdot-4y^1=-28x^6y \\ 3rd=\sum ^7_{i\mathop{=}2}(\frac{7!}{2!(7-2)!}x^{7-2}\cdot(-4y)^2=21\cdot x^5\cdot16y^2=336x^5y^2 \end{gathered}[/tex]
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