Break up by variable.

\(\displaystyle \frac{2^3}{4^2}= \frac{1}{2}\)

\(\displaystyle \frac{x^3}{x^4}= \frac{1}{x}\)

\(\displaystyle \frac{z^6}{z^2} = z^4\)

Therefore the simplified form becomes,

\(\displaystyle \frac{(2xz^2)^3}{(4zx^2)^2}=\frac{z^4}{2x}\).

C is computing the exponent, while A and B are equivalent due to properties of fractional exponents.

Remember that...

\(\displaystyle a^{\frac{b}{c}} = \sqrt[c]{a^b} = (\sqrt[c]{a})^b\)

\(\displaystyle 64^{\frac{4}{3}}=\) \(\displaystyle \sqrt[3]{64^4}=\) \(\displaystyle (\sqrt[3]{64})^{4}=\) \(\displaystyle 256\)

The easiest way to solve this is to simplify the fraction as much as possible. We can do this by factoring out the greatest common factor of the numerator and the denominator. In this case, the GCF is \(\displaystyle 7^7\). 

\(\displaystyle \frac{7^7\left ( 7^3-7^1\right)}{7^7\left ( 7^2-1\right)}\)

Now, we can cancel out the \(\displaystyle 7^7\) from the numerator and denominator and continue simplifying the expression.

\(\displaystyle \frac{7^7\left ( 7^3-7^1\right)}{7^7\left ( 7^2-1\right)}=\frac{7^3-7^1}{7^2-1}=\frac{343-7}{49-1}=\frac{336}{48}=7\)

\(\displaystyle (5x+2)^{2}\) is equivalent to \(\displaystyle (5x+2)(5x+2)\).

Using the FOIL method, you multiply the first number of each set \(\displaystyle 5x\cdot 5x=25x^{2}\), multiply the outer numbers of each set \(\displaystyle 5x\cdot 2=10x\), multiply the inner numbers of each set \(\displaystyle 2\cdot 5x=10x\), and multiply outer numbers of each set \(\displaystyle 2\cdot 2=4\).

Adding all these numbers together, you get \(\displaystyle 25x^{2}+10x+10x+4=25x^{2}+20x+4\). 

FOIL the first two terms:

\(\displaystyle (4x-9)(4x-9)=16x^2-36x-36x+81 = 16x^2-72x+81\)

Next, multiply this expression by the last term:

\(\displaystyle (16x^2-72x+81)(4x-9)=64x^3-144x^2-22x^2+648x+324x-729\)

Finally, combine the terms:

\(\displaystyle (4x-9)^3=64x^3-432x^2+972x-729\)

Plug in \(\displaystyle 2\) for \(\displaystyle q\) in the equation \(\displaystyle q(q-7)^{2}\)

That gives: \(\displaystyle 2(2-7)^{2}\)

Then solve the computation inside the parenthesis: \(\displaystyle 2(-5)^{2}\)

The answer should then be \(\displaystyle 50\)

Use FOIL and be mindful of exponent rules. Remember that when you multiply two terms with the same bases but different exponents, you will need to add the exponents together.

Remember to add exponents when two terms with like bases are being multiplied.

Use the FOIL method to simplify the following expression:

\(\displaystyle (x^3+2x^2)^2\)

Step 1: Expand the expression.

\(\displaystyle (x^3+2x^2)(x^3+2x^2)\)

Step 2: FOIL

First: \(\displaystyle x^3\cdot x^3 = x^6\)

Outside: \(\displaystyle x^3 \cdot 2x^2 =2x^5\)

Inside: \(\displaystyle 2x^2 \cdot x^3 = 2x^5\)

Last: \(\displaystyle 2x^2 \cdot 2x^2 = 4x^4\)

Step 2: Sum the products.

\(\displaystyle x^6+2x^5+2x^5+4x^4\)

\(\displaystyle x^6+4x^5+4x^4\)

Using FOIL on \(\displaystyle (a^3 + b^2c^2) \cdot (a^7 + bc^3)\), we see that:

First: \(\displaystyle a^3 \cdot a^7 = a^{10}\)

Outer: \(\displaystyle a^3 \cdot bc^3 = a^3bc^3\)

Inner: \(\displaystyle b^2c^2 \cdot a^7 = a^7b^2c^2\)

Last: \(\displaystyle b^2c^2 \cdot bc^3 = b^3c^5\)

Note that the middle terms are not additive: while they share common variables, they do not share matching exponents.

Thus, we have \(\displaystyle a^{10} + a^7b^2c^2 + a^3bc^3 + b^3c^5\). The arrangement goes by highest leading exponent, and alphabetically in the case of the last two terms.