How to Solve Cubic and Quartic Equations for Homework

[Brunno]

Homework Statement

Find the solution for :
X = \sqrt[3]{3+\sqrt{9+\frac{125}{27}}} \sqrt[]{-3+\sqrt{9+\frac{125}{27}}}

It doesn't appear above but is to subtract the first member by the second!

Any help will bevery welcome!
The answer is X= 1

Homework Equations

The Attempt at a Solution

 

[zgozvrm]

I think you also missed a "3"
I believe you want the second term to be a cube root also, correct?

If so, you should have:

\sqrt[3]{3+\sqrt{9 + \frac{125}{27}}} - \sqrt[3]{-3+\sqrt{9 + \frac{125}{27}}}

 

[zgozvrm]

Find the solution...

Just plug the numbers into a calculator!

 

[Brunno]

No.No calculator.Its a question that is to be developed and the answer must be x=1.So.did you get the answer?Give me some tip.

 

[zgozvrm]

Oh, that make things interesting...


I'm playing with the idea of setting

A = \sqrt{9 + \frac{125}{27}} + 3

and

B = \sqrt{9 + \frac{125}{27}} - 3

to give you

\sqrt[3]A - \sqrt[3]B

Then I'm trying to cube that and work from there...

\left( \sqrt[3]A - \sqrt[3]B \right)^3


I'm using the supposition that we already know that it's equal to 1, therefore cubing the statement won't change it's value.

 

[Brunno]

I'm using the supposition that we already know that it's equal to 1, therefore cubing the statement won't change it's value.

Why do you suppose that?Because in the question there's nothing sugesting the answer as x=1.Explain me please.

 

[zgozvrm]

Why do you suppose that?Because in the question there's nothing sugesting the answer as x=1.Explain me please.

First of all, you stated that the answer was 1.
Secondly, if we can prove that x^3 = 1[/tex] then x must be 1.

 

[zgozvrm]

Here's how to get started...


\sqrt[3]{3+\sqrt{9 + \frac{125}{27}}} - \sqrt[3]{-3+\sqrt{9 + \frac{125}{27}}}

can be rearranged to

\sqrt[3]{\sqrt{9 + \frac{125}{27}} + 3} - \sqrt[3]{\sqrt{9 + \frac{125}{27}} - 3}

and \LARGE {\sqrt[3]x = x^{1/3}}[/tex]<br /> <br /> so we have<br /> <br /> \left( \sqrt{9 + \frac{125}{27}} + 3 \right) ^{1/3} - \left( \sqrt{9 + \frac{125}{27}} - 3 \right) ^{1/3}<br /> <br /> Doing the math under the radical signs results in<br /> <br /> \left( \sqrt{\frac{368}{27}} + 3 \right) ^{1/3} - \left( \sqrt{\frac{368}{27}} - 3 \right) ^{1/3}<br /> <br /> which is equal to<br /> <br /> \left( \frac{\sqrt{368}}{\sqrt{27}} + 3 \right) ^{1/3} - \left( \frac{\sqrt{368}}{\sqrt{27}} - 3 \right) ^{1/3} = \left( \frac{\sqrt{368} \sqrt{27}}{\sqrt{27} \sqrt{27}} + 3 \right) ^{1/3} - \left( \frac{\sqrt{368} \sqrt{27}}{\sqrt{27} \sqrt{27}} - 3 \right) ^{1/3} = \left( \frac{\sqrt{9936}}{27} + 3 \right) ^{1/3} - \left( \frac{\sqrt{9936}}{27} - 3 \right) ^{1/3}<br /> <br /> = \left( \frac{\sqrt{9936} + 81}{27} \right) ^{1/3} - \left( \frac{\sqrt{9936} - 81}{27} \right) ^{1/3} = \frac{(\sqrt{9936} + 81)^{1/3} - (\sqrt{9936} - 81)^{1/3}}{3}<br /> <br /> So, we need to show that the numerator = 3.<br /> <br /> If you cube the whole term, you&#039;d have to show that the numerator is equal to 27.<br /> <br /> <br /> Cubing the numerator (I&#039;ll leave the messy work up to you) results in<br /> <br /> 162 - 45 \left[ (\sqrt{9936} + 81)^{1/3} - (\sqrt{9936} - 81)^{1/3} \right]<br /> <br /> Notice that the term inside the square brackets is the same term we cubed, so essentially what we have is<br /> <br /> X^3 = 162 -45X<br /> <br /> Solve this cubic equation and you get 3 answers; 1 real and 2 complex.<br /> You should find that the real answer is 3, therefore<br /> <br /> (\sqrt{9936} + 81)^{1/3} - (\sqrt{9936} - 81)^{1/3} = 3<br /> <br /> and the original statement is equal to 1.

 

[Brunno]

Thank you so much.I had no idea it had to use third degree equation,i don't even remember how to solve a 3° equation,but that's no problem.I will just work out this cube thing and see if i have the same answer.Thanks again!

 

[zgozvrm]

Here's a step-by-step on solving cubic equations:
http://www.1728.com/cubic2.htm

The site also has instructions for solving quartic equations (4th degree)