Sum of cubes question

[K^3]

This is the problem

= 4

I understand that I can substitute A for the first term and B for the second and have A + B = 4 then cube both sides and have a much simpler problem. However i was curious to see what would happen if you didn't substitute.



That is the result after cubing both sides(sorry it looks weird), and I am not complete sure where to go next. I'm pretty sure there are ways to factor the insides but I can't figure out what and how to do it.

 

[Deleted member 4993]

This is the problem

View attachment 33022 = 4

I understand that I can substitute A for the first term and B for the second and have A + B = 4 then cube both sides and have a much simpler problem. However i was curious to see what would happen if you didn't substitute.


View attachment 33023
That is the result after cubing both sides(sorry it looks weird), and I am not complete sure where to go next. I'm pretty sure there are ways to factor the insides but I can't figure out what and how to do it.

Are you using the formula:

(a + b)³ = a³ + b³ + 3a²b +3ab²

then

a³ = 14 + √x .......... b³ = 14 - √x...........and ............... a³ + b³ = 28 ............. continue

 

[topsquark]

This is the problem

View attachment 33022 = 4

I understand that I can substitute A for the first term and B for the second and have A + B = 4 then cube both sides and have a much simpler problem. However i was curious to see what would happen if you didn't substitute.


View attachment 33023
That is the result after cubing both sides(sorry it looks weird), and I am not complete sure where to go next. I'm pretty sure there are ways to factor the insides but I can't figure out what and how to do it.

Just to emphasize Subhotosh Khan's point you cubed what was under the radical but you didn't remove the radical! Remove the cube roots for the first and last terms.

-Dan

 

[JeffM]

It is great to have intellectual curiosity.

Here is another way to solve this problem.

[math]\sqrt[3]{14 + \sqrt{x}} + \sqrt[3]{14 - \sqrt{x}} = 4.\\ \text {Let } a = \sqrt{x}, \ b = \sqrt[3]{14 + a}, \text { and } c = \sqrt[3]{14 - a}.\\ \therefore \ b + c = 4 \implies b^3 + 3b^2c + 3bc^2 + c^3 = 64 \implies\\ 3bc(b + c) = 64 - (14 + a) - (14 - a) = 64 - 28 - a - (-a) = 36 \implies\\ bc(b + c) = 12 \implies 4bc = 12 \implies bc = 3 \implies\\ 3 = \sqrt[3]{14 + a} * \sqrt[3]{14 - a} = \sqrt[3]{196 - a^2} = \sqrt[3]{196 - x} \implies \\ 27 = 196 - x \implies x = 196 - 27 = 169.[/math]
Now let’s check.

[math]\sqrt[3]{14 + \sqrt{169}} + \sqrt[3]{14 - \sqrt{169}} = \sqrt[3]{14 + 13} + \sqrt[3]{14 - 13} = \\ \sqrt[3]{27} + \sqrt[3]{1} = 3 + 1 = 4. \ \checkmark[/math]
My point is that it is super-easy to make mistakes with radicals so it is worth considering whether substituting for all of them will simplify your life.

 

[K^3]

It is great to have intellectual curiosity.

Here is another way to solve this problem.

[math]\sqrt[3]{14 + \sqrt{x}} + \sqrt[3]{14 - \sqrt{x}} = 4.\\ \text {Let } a = \sqrt{x}, \ b = \sqrt[3]{14 + a}, \text { and } c = \sqrt[3]{14 - a}.\\ \therefore \ b + c = 4 \implies b^3 + 3b^2c + 3bc^2 + c^3 = 64 \implies\\ 3bc(b + c) = 64 - (14 + a) - (14 - a) = 64 - 28 - a - (-a) = 36 \implies\\ bc(b + c) = 12 \implies 4bc = 12 \implies bc = 3 \implies\\ 3 = \sqrt[3]{14 + a} * \sqrt[3]{14 - a} = \sqrt[3]{196 - a^2} = \sqrt[3]{196 - x} \implies \\ 27 = 196 - x \implies x = 196 - 27 = 169.[/math]
Now let’s check.

[math]\sqrt[3]{14 + \sqrt{169}} + \sqrt[3]{14 - \sqrt{169}} = \sqrt[3]{14 + 13} + \sqrt[3]{14 - 13} = \\ \sqrt[3]{27} + \sqrt[3]{1} = 3 + 1 = 4. \ \checkmark[/math]
My point is that it is super-easy to make mistakes with radicals so it is worth considering whether substituting for all of them will simplify your life.

Thank you for taking the time to solve the problem. However, I must not have made myself clear. I did solve the problem with a substitution methodology. I wanted to see if I could multiply out the cubes long hand. Indeed I was successful. Unfortunately, I cannot figure out how to simply the equation any further to solve for X(from the second attatchment). I ran the long-hand calculation through Demos.com and it verified that the equation and answer were correct. It did not provide a means to simplify. I think I would really benefit from seeing how this long-hand version approach would work out. Any thoughts would be great, but thank you again for your time. -Kurt

 

[K^3]

Just to emphasize Subhotosh Khan's point you cubed what was under the radical but you didn't remove the radical! Remove the cube roots for the first and last terms.

-Dan

I was of the understanding that if Then I would have to do something to the inside of the radical to remove the radical.

 

[Cubist]

IF we're only considering the real value solution of the cube root function (also real valued input to the cube root, therefore x≥0) then...

\(\displaystyle \left(\sqrt[3]{14+\sqrt{x}}\right)^3\)
\(\displaystyle = 14 + \sqrt{x}\)

It seems very painful to go down this route (which has equivalent value)...
\(\displaystyle \left(\sqrt[3]{14+\sqrt{x}}\right)^3\)
\(\displaystyle = \sqrt[3]{\left(14+\sqrt{x}\right)^3}\)
\(\displaystyle = \sqrt[3]{2744 + 588\sqrt{x} + 42x + x\sqrt{x}}\)

I'm not surprised that a computer algebra system doesn't spot that [imath]2744 + 588\sqrt{x} + 42x + x\sqrt{x}[/imath] is a perfect cube (which is the way of reversing out of this mess). Although, Wolfram(click) manages to spot it.

 

[JeffM]

[math]\sqrt[3]{14 + \sqrt{x}} + \sqrt[3]{14 - \sqrt{x}} = 4 \implies \left (\sqrt[3]{14 + \sqrt{x}} + \sqrt[3]{14 - \sqrt{x}} \right )^3 = 4^3 \implies \\ 64 = \left (\sqrt[3]{14 + \sqrt{x}} \right )^3 + 3 \left ( \sqrt[3]{14 + \sqrt{x}} \right )^2 * \sqrt[3]{14 - \sqrt{x}} + 3\sqrt[3]{14 + \sqrt{x}} * \left ( \sqrt[3]{14 - \sqrt{x}} \right )^2 + \left ( \sqrt[3]{14 - \sqrt{x}} \right )^3.\\ \therefore 64 = 14 + \sqrt{x} + 3 * \sqrt[3]{196 - (\sqrt{x})^2} * \left ( \sqrt[3]{14 + \sqrt{x}} + \sqrt[3]{14 - \sqrt{x}} \right ) + 14 - \sqrt{x}.[/math]
But [imath]\sqrt[3]{14 + \sqrt{x}} + \sqrt[3]{14 - \sqrt{x}} = 4[/imath] by hypothesis. So we can simplify

[math]64 = 28 + 3 * 4 \sqrt[3]{196 - x} \implies 36 = 12 \sqrt[3]{196 - x} \implies 3 = \sqrt[3]{196 - x} \implies \\ 27 = 196 - x \implies x = 196 - 27 = 169.[/math]
Why would you do all that? The opportunity for error is huge. And if you did not see the opportunity to replace a sum of cube roots with four, you go on and on for quite a while.