Question

expand each binomial.
$(3m - 4p)^3$
$\bigcirc\\ 27m^3 + 108m^2p - 144mp^2 + 64p^3$
$\bigcirc\\ 27m^3 - 108m^2p + 144mp^2 - 64p^3$
$\bigcirc\\ 27m^3 - 64p^3$
$\bigcirc\\ 27m^3 - 108m^2p^3 + 144mp^2 - 64p$

question 7
1 pts
find the indicated term of each expansion.
sixth term of $(x + 4y)^6$
$\bigcirc\\ 4096xy^5$
$\bigcirc\\ 6144xy^5$
$\bigcirc\\ 6144x^5y^3$
$\bigcirc\\ 4096x^6y^6$

Explanation:

First Question: Expand \(\boldsymbol{(3m - 4p)^3}\)

Step1: Recall the binomial cube formula

The formula for \((a - b)^3\) is \(a^3 - 3a^2b + 3ab^2 - b^3\). Here, \(a = 3m\) and \(b = 4p\).

Step2: Calculate each term

• For \(a^3\): \((3m)^3 = 27m^3\)
• For \(-3a^2b\): \(-3\times(3m)^2\times(4p)= -3\times9m^2\times4p = -108m^2p\)
• For \(3ab^2\): \(3\times(3m)\times(4p)^2 = 3\times3m\times16p^2 = 144mp^2\)
• For \(-b^3\): \(-(4p)^3 = -64p^3\)

Step3: Combine the terms

Putting them together, we get \(27m^3 - 108m^2p + 144mp^2 - 64p^3\).

Step1: Recall the binomial theorem term formula

The \(k\)-th term (starting from \(k = 1\)) in the expansion of \((a + b)^n\) is given by \(T_{k}=\binom{n}{k - 1}a^{n-(k - 1)}b^{k - 1}\). Here, \(n = 6\), \(k = 6\), \(a = x\), \(b = 4y\).

Step2: Calculate the binomial coefficient

First, find \(\binom{6}{6 - 1}=\binom{6}{5}\). We know that \(\binom{n}{r}=\frac{n!}{r!(n - r)!}\), so \(\binom{6}{5}=\frac{6!}{5!(6 - 5)!}=\frac{6!}{5!1!}=\frac{6\times5!}{5!×1}=6\).

Step3: Calculate the powers of \(a\) and \(b\)

For \(a^{n-(k - 1)}\): \(a^{6-(6 - 1)}=a^{1}=x\).
For \(b^{k - 1}\): \(b^{6 - 1}=(4y)^{5}=4^{5}y^{5}=1024y^{5}\).

Step4: Multiply the coefficient, \(a\) term, and \(b\) term

Now, \(T_{6}=\binom{6}{5}\times x\times(4y)^{5}=6\times x\times1024y^{5}=6144xy^{5}\).

Answer:

\(27m^3 - 108m^2p + 144mp^2 - 64p^3\) (the second option)

Second Question: Find the sixth term of \(\boldsymbol{(x + 4y)^6}\)