To solve this equation, we have to factor our radicals.  We do this by finding numbers that multiply to give us the number within the radical. 

\(\displaystyle \sqrt{80} = \sqrt{20}\sqrt{4}\)

\(\displaystyle \sqrt{20} = \sqrt{5}\sqrt{4}\)

Add them together:

\(\displaystyle \sqrt5\sqrt4\sqrt4 + \sqrt5\sqrt4\)

4 is a perfect square, so we can find the root:

\(\displaystyle (2)(2)\sqrt5 + 2\sqrt5\)

\(\displaystyle 4\sqrt5 + 2\sqrt5 = 6\sqrt5\)

Since both have the same radical, we can combine them:

\(\displaystyle 4\sqrt5 + 2\sqrt5 = 6\sqrt5\)

When simplifying, you should always be on the lookout for like terms. While it might not look like there are like terms in \(\displaystyle f(x)\), there are -- we just have to be able to rewrite it to see.

\(\displaystyle f(x)=x^{2}-3\sqrt{4x}+4x^{\frac{1}{3}}-x-7x^{\frac{1}{2}}+5\sqrt[3]{x}\)

\(\displaystyle f(x)=x^{2}-3(4x)^{\frac{1}{2}}+4x^{\frac{1}{3}}-x-7x^{\frac{1}{2}}+5x^{\frac{1}{3}}\)

Before we start combining terms, though, let's look a little more closely at this part:

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

We need to "distribute" that exponent to everything in the parentheses, like so:

\(\displaystyle -3\cdot 4^{\frac{1}{2}}\cdot x^{\frac{1}{2}}\)

But 4 to the one-half power is just the square root of 4, or 2.

\(\displaystyle -3\cdot 2\cdot x^{\frac{1}{2}}=-6x^{\frac{1}{2}}\)

Okay, now let's see our equation.

\(\displaystyle f(x)=x^{2}-6x^{\frac{1}{2}}+4x^{\frac{1}{3}}-x-7x^{\frac{1}{2}}+5x^{\frac{1}{3}}\)

We need to start combining like terms. Take the terms that include x to the one-half power first.

\(\displaystyle f(x)=x^{2}\mathbf{-6x^{\frac{1}{2}}}+4x^{\frac{1}{3}}-x\mathbf{-7x^{\frac{1}{2}}}+5x^{\frac{1}{3}}=x^{2}\mathbf{-13x^{\frac{1}{2}}}+4x^{\frac{1}{3}}-x+5x^{\frac{1}{3}}\)

Now take the terms that have x to the one-third power.

\(\displaystyle f(x)=x^{2}-13x^{\frac{1}{2}}\mathbf{+4x^{\frac{1}{3}}}-x\mathbf{+5x^{\frac{1}{3}}}=x^{2}-13x^{\frac{1}{2}}\mathbf{+9x^{\frac{1}{3}}}-x\)

All that's left is to write them in order of descending exponents, then convert the fractional exponents into radicals (since that's what our answer choices look like).

\(\displaystyle f(x)=x^{2}-x-13x^{\frac{1}{2}}+9x^{\frac{1}{3}}\)

\(\displaystyle f(x)=x^{2}-x-13\sqrt{x}+9\sqrt[3]{x}\)

We can simplify both radicals:

 \(\displaystyle \sqrt{16} =4\)            and              \(\displaystyle \sqrt{8}=\sqrt{4\times 2} = 2\sqrt{2}\)

Plug in the simplified radicals into the equation:

\(\displaystyle 2(4) - 4(2\sqrt{2})\)

which leaves us with: 

\(\displaystyle 8- 8\sqrt{2}\)

Because these are not like terms, we cannot simplify this further.

Only the first two radicals can be simplified: 

\(\displaystyle \sqrt{27}=\sqrt{9\times 3}=3\sqrt{3}\)      and    \(\displaystyle \sqrt{9}=3\)

Plug in the simplified radicals into the equation: 

\(\displaystyle -8(3\sqrt{3})-11(3)+2\sqrt{3}\)

\(\displaystyle -24\sqrt{3}-33+2\sqrt{3}\)

We can now simplify the equation by combining the like terms: 

\(\displaystyle -22\sqrt{3}-33\)

We can simplify the first and third radicals:

\(\displaystyle \sqrt{28} =\sqrt{7\times 4} =2\sqrt{7}\)       and   \(\displaystyle \sqrt{63} =\sqrt{9\times 7 } =3\sqrt{7}\)

Plug in the simplified radicals into the equation:

\(\displaystyle 5(2\sqrt{7}) + \sqrt{7} -7(3\sqrt{7})\)

\(\displaystyle 10\sqrt{7} + \sqrt{7 } -21\sqrt{7}\)

Combine the like terms: 

\(\displaystyle -10\sqrt{7}\)

 \(\displaystyle Think \: of\; x = \sqrt7\)

\(\displaystyle 3x + 4x = 7x\)

 \(\displaystyle Therefore, \: 3\sqrt{7} + 4\sqrt{7} = 7\sqrt7\)

The third root of \(\displaystyle 27\) is \(\displaystyle 3\)

\(\displaystyle \left ( 3\times 3\times 3 =27\right )\)

and when added to the square root of 64, which is 8, you should get 11.  

When adding and subtracting radicals, make the sure radicand or inside the square root are the same.

If they are the same, just add the numbers in front of the radical.

Since they are not the same, the answer is just the problem stated. 

\(\displaystyle {}\sqrt{2}+\sqrt{3}\)

When adding and subtracting radicals, make the sure radicand or inside the square root are the same.

If they are the same, just add the numbers in front of the radical.

Since they are not the same, the answer is just the problem stated. 

\(\displaystyle \sqrt{7}-\sqrt{4}\)

When adding and subtracting radicals, make the sure radicand or inside the square root are the same.

If they are the same, just add the numbers in front of the radical.

Looking carefully at the radicand we will notice that each radicand is a perfect sqare. This means we are able to reduce the radical into an integer.

 \(\displaystyle 9, 4,\) and \(\displaystyle 1\) are all sqaure numbers so instead we have a simple algebraic problem: 

\(\displaystyle \sqrt9+\sqrt4 +\sqrt1=\sqrt{3\cdot 3}+\sqrt{2\cdot 2}+\sqrt{1\cdot 1}=3+2+1\) which the answer is \(\displaystyle 6\).