solve the equation...

[poiuy]

\(\displaystyle \sqrt[3]{x+1}+\sqrt[3]{3x+1}=\sqrt[3]{x-1}\)

How can this be done?

 

[Ryan Rigdon]

raise both sides to the third power and then you get


(x+1) + (3x+1) = (x-1)

4x + 2 = x - 1

3x = -3

x = -1

when you check it and plus it back into original equation you should get (-2)^(1/3) = (-2)^(1/3).

 

[pappus]

raise both sides to the third power and then you get


(x+1) + (3x+1) = (x-1)

4x + 2 = x - 1 <--- :shock:

3x = -3

x = -1

when you check it and plus it back into original equation you should get (-2)^(1/3) = (-2)^(1/3).

... you are a lucky guy: You did this question wrong - but your solution is correct!

\(\displaystyle \sqrt[3]{x+1}+\sqrt[3]{3x+1}=\sqrt[3]{x-1}\)

How can this be done?

1. As Ryan Rigdon suggested raise bot sides to the 3rd power:

\(\displaystyle \sqrt[3]{x+1}+\sqrt[3]{3x+1}=\sqrt[3]{x-1}\)

\(\displaystyle (x+1) + 3\sqrt[3]{(x+1)^2 \cdot (3x+1)} + 3\sqrt[3]{(x+1) \cdot (3x+1)^2} + (3x+1)=x-1\)

\(\displaystyle 3\sqrt[3]{(x+1)^2 \cdot (3x+1)} + 3\sqrt[3]{(x+1) \cdot (3x+1)^2} + 3x+3=0\)

Divide through both sides by 3:

\(\displaystyle \sqrt[3]{(x+1)^2 \cdot (3x+1)} + \sqrt[3]{(x+1) \cdot (3x+1)^2} + (x+1)=0\)

Now factor out \(\displaystyle \sqrt[3]{x+1}\) :

\(\displaystyle \sqrt[3]{x+1} \cdot \left( \sqrt[3]{(x+1) \cdot (3x+1)} + \sqrt[3]{ (3x+1)^2} + \sqrt[3]{(x+1)^2} \right)=0\)

A product can only be zero if at least one factor equals zero:

\(\displaystyle \sqrt[3]{x+1} = 0~\vee~\sqrt[3]{(x+1) \cdot (3x+1)} + \sqrt[3]{ (3x+1)^2} + \sqrt[3]{(x+1)^2} = 0\)

From the first factor you'll get: \(\displaystyle x = -1\)

And now you have to prove that the 2nd factor can't be zero. Good luck!

 

[poiuy]

\(\displaystyle \sqrt[3]{(x+1)\cdot(3x+1)}+\sqrt[3]{(3x+1)^{2}}+\sqrt[3]{(x+1)^{2}}=0\)

I'm sorry that it writing 10 days after you reply but ...
I can't prove that it has no solutions toring:-/
I tried again to raise the three powers but none of that:-/,
Whatcan I do?

 

[Deleted member 4993]

What is the domain of "x"?

 

[poiuy]

All real numbers

 

[Deleted member 4993]

\(\displaystyle \sqrt[3]{(x+1)\cdot(3x+1)} +\sqrt[3]{(3x+1)^{2}}+\sqrt[3]{(x+1)^{2}}=0\)

I'm sorry that it writing 10 days after you reply but ...
I can't prove that it has no solutions toring:-/
I tried again to raise the three powers but none of that:-/,
Whatcan I do?

What is the minimum value of

\(\displaystyle \sqrt[3]{(x+1)\cdot(3x+1)}\)

and at what value of 'x' does that minimum occur?

 

[poiuy]

ok,I see that I am doing something wrong:-D
I thought that I count the value of the root of the quadratic function and then came to
\(\displaystyle \sqrt[3]{-\frac{2}{3}}\) for \(\displaystyle x=-\frac{1}{3}\)
What i'm doing wrong?

 

[pappus]

\(\displaystyle \sqrt[3]{(x+1)\cdot(3x+1)}+\sqrt[3]{(3x+1)^{2}}+\sqrt[3]{(x+1)^{2}}=0\)

I'm sorry that it writing 10 days after you reply but ...
I can't prove that it has no solutions toring:-/
I tried again to raise the three powers but none of that:-/,
Whatcan I do?

1. The last 2 summands are greater or equal to zero because the argument of the cube root is squared. Thus the 1st summand must be negative to get a final result of zero.

2. You easily can prove that the minimum of

\(\displaystyle (x+1)(3x+1)\) is at \(\displaystyle x = -\frac23\) which yields \(\displaystyle \sqrt[3]{\frac13 \cdot (-1)} = \sqrt[3]{-\frac13}\)

3. At \(\displaystyle x = -\frac23\) the positive summands are:

\(\displaystyle \sqrt[3]{1}+\sqrt[3]{\frac19}\)

So this sum is greater than 1.

4. Since \(\displaystyle -1 < \sqrt[3]{-\frac13} < 0\) and the the last 2 summands beeing greater than 1 the whole term can't never be zero.