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Question

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allieeslabae · Answer

attachment

johnweldon1993 · Answer

$$\large 3\sqrt[5]{(x + 2)^3} + 3 = 27$$

Work backwards and solve for 'x'

Subtract 3 from both sides

$$\large 3\sqrt[5]{(x + 2)^3} = 24$$

Divide both sides by 3

$$\large \sqrt[5]{(x + 2)^3} = 8$$
Raise both sides to the 5th power

$$\large (x + 2)^3 = 8^5$$

Lets just write that as 

$$\large (x + 2) ^ 3 = 32768$$

Take the cubed root of both sides now

$$\large x + 2 = 32$$

And I think you know the rest :P

jhonyy9 · Answer

little simple than 

(x+2)^3 = 8^5  you know that 8=2^3 so than you can rewrite 8^5 = (2^3)^5 = (2^5)^3

(x+2)^3 = (2^5)^3   make cuberoot from both sides and will get 

x+2=2^5

x+2=32 and now you can ending it sure easy 

hope this more easy way will be understand it sure right easy 

good luck

bye

johnweldon1993 · Answer

$$\large 3\sqrt[3]{2a} - 6\sqrt[3]{2a}$$

Notice how both terms have $\large \sqrt[3]{2a}$    lets make that = x *just for simplicity*

So lets re-write this as

$$\large 3x - 6x$$

What would that be?

allieeslabae · Answer

Okay so we're combining like terms. and we have -3x

johnweldon1993 · Answer

Correct

So lets just re-sub in what we made 'x' equal

$$\large -3\sqrt[3]{2a}$$

Now...this can still be simplified...remember the rule of exponents

$$\Large \sqrt[b]{x^a} = x^{\frac{a}{b}}$$

How can we apply that to this problem?

allieeslabae · Answer

Okay so we make it -3^3(2a) ? Is that in the right order.

johnweldon1993 · Answer

Not quite

So for now...ignore that -3 out in front

$$\large \sqrt[3]{2a}$$

If we apply the fact that $\large \sqrt[b]{x^a} = x^{\frac{a}{b}}$

We would have

$$\large (2a)^{\frac{1}{3}}$$

Right?

allieeslabae · Answer

Okay, i'm listening.

johnweldon1993 · Answer

Well that's literally it...now we just bring back in the -3 we ignored before

$$\large -3(2a)^{\frac{1}{3}}$$ would be your final answer

allieeslabae · Answer

Okay wait, you know how the 3 was by the sqrt what does that mean? How did it become cubed?

johnweldon1993 · Answer

$\large \sqrt{x}$ you know that is seen as "the square root of 'x' right?

johnweldon1993 · Answer

Well to get rid of that "eliminate it" we do the inverse operation of taking a "square root"

This operation is raising to the 2nd power "squaring it"

johnweldon1993 · Answer

Same thing for this problem

$\large \sqrt[3]{x}$ is seen as the "cubed root of x"

To get rid of it ...we raise this to the 3rd power *cube it*

johnweldon1993 · Answer

So all in all...

$$\large (\sqrt[3]{x})^3 = x$$

allieeslabae · Answer

Ahh Okay, thanks smarty.

johnweldon1993 · Answer

Well of course =[p

anonymous · Answer

Question3:
A radical can be expressed as an expression with a fractional exponent by following the convention. Rewriting radicals using fractional exponents can be useful can be useful in simplifying some radical expression. When working with fractional exponents, remember that fractional exponents are subject to all of the same rules as other exponents when they appear in algebraic expressions.$$\sqrt[n]{x ^{m}}= x \frac{ m }{ n }$$

allieeslabae · Answer

Hey Thanks @Austin1617 I have no clue why that was listed, I have it. Too kind. :)

anonymous · Answer

Okay just trying to help.

allieeslabae · Answer

I know, and thank you so much for that. Look out for anymore in the question! :)