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#### mathmaniac

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- Mar 4, 2013

- 188

1)\(\displaystyle \sqrt{A}+\sqrt{B}+\sqrt{C}=D\)

2)A

^{1/3 }+ B

^{1/3 }= C

Or how else can it be solved?

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- Mar 4, 2013

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1)\(\displaystyle \sqrt{A}+\sqrt{B}+\sqrt{C}=D\)

2)A

Or how else can it be solved?

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- #3

- Mar 4, 2013

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1)\(\displaystyle \sqrt{A}+\sqrt{B}+\sqrt{C}+\sqrt{D}=E\)

Please show how to solve modified 1) and 2),i.e, make them polynomial

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2.) \(\displaystyle \sqrt[3]{A}+\sqrt[3]{B}=C\)

Cube both sides to get:

\(\displaystyle A+3\sqrt[3]{A^2B}+3\sqrt[3]{AB^2}+B=C^3\)

\(\displaystyle 3\sqrt[3]{AB}\left(\sqrt[3]{A}+\sqrt[3]{B} \right)=C^3-(A+B)\)

The original equation allows us to simplify as:

\(\displaystyle 3C\sqrt[3]{AB}=C^3-(A+B)\)

Cube again:

\(\displaystyle 27ABC^3=C^9-3C^6(A+B)+3C^3(A+B)^2+(A+B)^3\)

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- Mar 4, 2013

- 188

Please give an answer for 1) too

I tried and failed

General Question:Can this be done for any eq with rational powers?

- Jan 29, 2012

- 1,151

It should be obvious that the answer is yes.

Please give an answer for 1) too

I tried and failed

General Question:Can this be done for any eq with rational powers?

As for your previous question please show

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- Mar 4, 2013

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\(\displaystyle 2\sqrt{AB}+2\sqrt{AC}+2\sqrt{AD}+2\sqrt{BC}+2\sqrt{BD}+2\sqrt{CD}....\)

I can take common \(\displaystyle 2\sqrt{A}\)and sub for \(\displaystyle \sqrt{B}+\sqrt{C}+\sqrt{D}\)

Whatever I tried to do then didn't help....

Whoever's going to reply only needs to show how to reduce the number of rooted terms,make it less than 4...

I think there's some trick or maybe just my problem.

Please help

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- Feb 7, 2012

- 2,725

Take one term across to the other side before squaring: $\sqrt A + \sqrt B = D - \sqrt C$.

\(\displaystyle 2\sqrt{AB}+2\sqrt{AC}+2\sqrt{AD}+2\sqrt{BC}+2\sqrt{BD}+2\sqrt{CD}....\)

I can take common \(\displaystyle 2\sqrt{A}\)and sub for \(\displaystyle \sqrt{B}+\sqrt{C}+\sqrt{D}\)

Whatever I tried to do then didn't help....

Whoever's going to reply only needs to show how to reduce the number of rooted terms,make it less than 4...

I think there's some trick or maybe just my problem.

Please help

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- #9

- Mar 4, 2013

- 188

I meant the modified 1Take one term across to the other side before squaring: $\sqrt A + \sqrt B = D - \sqrt C$.

See my edit.

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- Feb 7, 2012

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[I should have read the whole thread – I didn't see the modification.]I meant the modified 1

See my edit.

Try an inductive method. Start with the equation $\sqrt{A}+\sqrt{B}+\sqrt{C}=F$. That only has three roots, so you (presumably) know how to remove them to get a polynomial equation in powers of $A$, $B$, $C$ and $F$. Now substitute $F = E - \sqrt D$ into that equation. You will then have an equation in which the only radical is $\sqrt D$. So collect all the terms with $\sqrt D$ onto one side of that equation and square again.

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- #11

- Mar 5, 2012

- 8,885

Alternatively, substitute $A=x^6, B=y^6, C=z^6$ with $x,y,z \ge 0$.

1)\(\displaystyle \sqrt{A}+\sqrt{B}+\sqrt{C}=D\)

2)A^{1/3 }+ B^{1/3 }= C

Or how else can it be solved?

Then you get:

1) \(\displaystyle x^3+y^3+z^3=D\)

2) \(\displaystyle x^2+ y^2= z^6\)

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- #12

- Mar 4, 2013

- 188

No,I was thinking about A,B,C...representing polynomial functions on a variable (say x) but RHS a constant.There you can't do these kind of techniques when your purpose is to solve for x.Alternatively, substitute $A=x^6, B=y^6, C=z^6$ with $x,y,z \ge 0$.

Then you get:

1) \(\displaystyle x^3+y^3+z^3=D\)

2) \(\displaystyle x^2+ y^2= z^6\)

But nice (funny?) answer,for my less informative post.

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- #13

- Mar 4, 2013

- 188

Very very nice trick,impressive!!![I should have read the whole thread – I didn't see the modification.]

Try an inductive method. Start with the equation $\sqrt{A}+\sqrt{B}+\sqrt{C}=F$. That only has three roots, so you (presumably) know how to remove them to get a polynomial equation in powers of $A$, $B$, $C$ and $F$. Now substitute $F = E - \sqrt D$ into that equation. You will then have an equation in which the only radical is $\sqrt D$. So collect all the terms with $\sqrt D$ onto one side of that equation and square again.

using this,i can make any eq with only degree as 1/2 a polynomial no matter,how many terms there are...

I'll be soon bringing another question about rational degrees(maybe) and I think the general question can also be solved by induction.Right???

Let me see.

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