Help please!!!!!!!!!

**Ruler of Hell** · June 28th 2006, 07:27 AM

Hi, I'm 13 years old and I'm from India. I just found out about the forum. Looking a some posts over here, I see that you guys are uber geniuses and I thought that you could provide me a helping hand.

1) If x+y+z=0, show that x³+y³+z³ = 3xyz.
2) If a+b+c=6, a²+b²+c²=14 and a³+b³+c³=26, then find the value of abc.
3) Find x³-1/x³ if x-1/x=a
4) Find x³+1/x³ if x+1/x=4
5) Find x³+1/x³ if x+1/x=√3 (That last sign is a root sign)
6) If x+1/x=p then show that x³+1/x³=p(p²-3)

These are only 6 sums out of about 11 or so. I don't like posting so many either but I can't understand anything. Like in say the 3rd sum, I've been taught how do you solve if the ³ was replaced by a ².

P.S. I'm yet learning the code the PHP or whatever code, so please forgive me for the way I'm posting.

Edit: I thought it'd be better if I posted in a more decent manner now. Thank you guys!

**ThePerfectHacker** · June 28th 2006, 07:47 AM

Originally Posted by Ruler of Hell

1) If x+y+z=0, show that x³+y³+z³ = 3xyz.

Thus,
$x+y=-z$
Cube both sides,
$(x+y)^3=(-z)^3$
Expand the left side and redo the right,
$x^3+3x^2y+3xy^2+y^3=-z^3$
Thus,
$(x^3+y^3+z^3)+3xy(x+y)=0$
But,
$x+y=-z$
Thus,
$(x^3+y^3+z^3)+3xy(-z)=0$
Thus,
$(x^3+y^3+z^3)-3xyz=0$
Finally,
$x^3+y^3+z^3=3xyz$

**Ruler of Hell** · June 28th 2006, 07:53 AM

I so love you! Seriously man you rock!!!!!!!!! Thanks. One sum is better than nothing. My father taught me this sum but he never showed the sum whereby you have to replace the (x+y) with -z. Listen, can you please come on MSN or Gmail? Tell me your id or anything! Please! I need lots of help! *Writes the sum down*
Btw, my mother has told you thanks for helping me!
*Hopes that ThePerfectHacker will come for chatting*

**CaptainBlack** · June 28th 2006, 08:46 AM

Originally Posted by Ruler of Hell

3) Find x³-1/x³ if x-1/x=a

$a^3=(x-1/x)^3=x^3-3x^2/x+3x/x^2-1/x^3$

$ =x^3-1/x^3-3(x-1/x) $

$ =x^3-1/x^3 -3a $

So:

$ x^3-1/x^3=a^3+3a $

RonL

You should check this carefully, I'm not on form in the accuracy
department at the moment.

**Ruler of Hell** · June 28th 2006, 10:32 AM

Thanks, that helped me solve 4 sums! Yay! I'll post some more now...

Okay, I'm stuck on this part of a sum.
x³+1/x³ +3√3 = √3 ³

Now what?? Here's the actual sum.

If x+1/x = √3
Find x³+1/x³

Another sum x+1/x=p
Show that x³+1/x³=p(p²+3)
I don't see the reason for writing it in a weird manner. Can I write this at the last step:-
p³+3p and then change it to
p(p²+3)

So this is where I'm stuck. Thanks again. This forum rocks!!!!

**Soroban** · June 28th 2006, 01:45 PM

Hello, Ruler of Hell!

$\text{5) Find }x^3 + \frac{1}{x^3} \text{ if }x + \frac{1}{x}\:=\:\sqrt{3}$

We are given: . $x + \frac{1}{x} \:=\:\sqrt{3}$ .*

Cube both sides: . $\left(x + \frac{1}{x}\right)^3\:=\<img src=$ \sqrt{3})^3" alt="\left(x + \frac{1}{x}\right)^3\:=\

\sqrt{3})^3" />

And we have: . $x^3 + 3x + \frac{3}{x} + \frac{1}{x^3} \;= \; 3\sqrt{3}$

Regroup the terms: . $\left(x^3 + \frac{1}{x^3}\right) + 3\underbrace{\left(x + \frac{1}{x}\right)} \;= \; 3\sqrt{3}$
. . . . . . . . . . . . . . . . . . . . . . . . This is $\sqrt{3}$ *

And we have: . $\left(x^3 + \frac{1}{x^3}\right) + 3\sqrt{3} \;= \;3\sqrt{3}$

Therefore: . $x^3 + \frac{1}{x^3} \;= \;0$

$\text{6) If }x + \frac{1}{x}\,=\,p\text{, then show that: }x^3 + \frac{1}{x^3}\:=\<img src=$ (p^2 - 3)" alt="\text{6) If }x + \frac{1}{x}\,=\,p\text{, then show that: }x^3 + \frac{1}{x^3}\:=\

(p^2 - 3)" />

Use the same procedure . . .

We are given: . $x + \frac{1}{x} \;= \;p$

Cube: . $x^3 + 3x + \frac{3}{x} + \frac{1}{x^3} \;= \; p^3$

Regroup: . $\left(x^3 + \frac{1}{x^3}\right) + 3\underbrace{\left(x + \frac{1}{x}\right)}\;=\;p^3$
. . . . . . . . . . . . . . . . . This is $p$

And we have: . $\left(x^3 + \frac{1}{x^3}\right) + 3p \;= \;p^3$

Therefore: . $x^3 + \frac{1}{x^3} \;= \; p^3 - 3p \;= \;p(p^2 - 3)$

**Ruler of Hell** · June 28th 2006, 06:51 PM

Thanks a lot Soroban. I think I have only another sum or two which I don't know and then I can submit the worksheet!

$ \text{1) If }x^2 + \frac{1}{25x^2} = \frac{43}{5} \text{ then find the value of }x^3 + \frac{1}{125x^3} $

Find a³ + b³ + c³ - 3abc
......ab+bc+ac -a² - b² - c²

When, a=-5 b=-6 c=10

That whole thing is a fraction, it wouldn't let me type it as a code. Some [LaTeX Error: Couldn't copy temporary file]

My friend suggested something like multiplying it by -1. Please help!

Those are the only two sums I can't figure out. Both of them I've gotten half way through but I can't get a step or two in between. *Gives everyone a bear hug for helping* Oh, and yet again I wish to say Thank You! My grandmother is also happy that I'm seeking help from you guys and my mother will be really delighted once I tell her about you guys. Thanks!

**Soroban** · June 28th 2006, 07:34 PM

Hello, Ruler of Hell!

$\text{1) If }x^2 + \frac{1}{25x^2} = \frac{43}{5} \text{ then find the value of }x^3 + \frac{1}{125x^3} $

This one required Olympic-level gymnastics . . .

We have: . $x^2 + \frac{1}{25x^2}\:=\;\frac{43}{5}$

Add $\frac{2}{5}$ to both sides: . $x^2 + \frac{2}{5} + \frac{1}{25x^2} \;=\;\frac{43}{5} + \frac{2}{5}\;=\;9$

. . and we have: . $\left(x + \frac{1}{5x}\right)^2\;=\;9\quad\Rightarrow\quad \boxed{x + \frac{1}{5x}\;=\;3}$

Cube both sides: . $x^3 + \frac{3x}{5} + \frac{3}{25x} + \frac{1}{125x^3} \;= \;27$

Regroup: . $\left(x^3 + \frac{1}{125x^3}\right) +\frac{3}{5}\underbrace{\left(x + \frac{1}{5x}\right)} \;= \;27$
. . . . . . . . . . . . . . . . . . . This is $3$

. . so we have: . $\left(x^3 + \frac{1}{125x^3}\right) + \frac{3}{5}(3)\;=\;27\quad\Rightarrow\quad x^3 + \frac{1}{125x^3} \;= \;27 - \frac{9}{5}$

Therefore: . $\boxed{x^3 + \frac{1}{125x^3} \;= \;\frac{126}{5}}$

**Ruler of Hell** · June 29th 2006, 08:35 AM

Woohoo yeah!

Now one last sum to go! I love you guys!

I've already posted it previously! You guys are so cool! Now I never need to fear Math again. Thanks for teaching me!!!! *Waits for somebody to solve the last sum....

*

**Soroban** · June 29th 2006, 08:40 AM

Hello, Ruler of Hell!

I finally got #2 . . .
I assume there is a typo . . . The sum of cubes should be 36, right?

2) If $\begin{Bmatrix} [1]\;a+b+c=6 \\ [2]\;a^2+b^2+c^2=14 \\ [3]\;a^3+b^3+c^3=36\end{Bmatrix}$ , then find the value of $abc.$

From $[1]\;a + b + c \,=\,6$ , we have: . $\begin{Bmatrix}a + b = 6 - c \\ b + c = 6 - a \\ a + c = 6 - b\end{Bmatrix}$ (a)

Square $[1]:\;\;(a + b + c)^2\:=\:6^2$

We have: . $\underbrace{a^2 + b^2 + c^2} + \:2(ab + bc + ac) \:=\:36$
From $[2]$ , this is $14$

Then: . $14 + 2(ab + bc + ac) \:= \:36\quad\Rightarrow\quad ab + bc + ac\:=\:11$ (b)

Cube $[1]:\;\;(a + b + c)^3\:=\:6^3$

We have: . $\underbrace{a^3 + b^3 + c^3} + \:3(a^2b + ab^2 + b^2c + bc^2 + a^2c + ac^2) + 6abc$ $\;= \;216$
From $[3]$ , this is $36$

Then: . $36 + 3(a^2b + ab^2 + b^2c + bc^2 + a^2c + ac^2) + 6abc$ $\;=\;216$

. . and we have: . $a^2b + ab^2 + b^2c + bc^2 + a^2c + ac^2 + 2abc\;=\;60$

Factor: . . . $ab\underbrace{(a + b)} +\, bc\underbrace{(b + c)} + \,ac\underbrace{(a + c)} + \,2abc\;=\;60$
From (a): . $ab(6 - c) + bc(6 - a) + ac(6 - b) + 2abc \;= \;60$

We have: . $6ab - abc + 6bc - abc + 6ac - abc + 2abc \;= \;60$

Simplify: . $6\underbrace{(ab + bc + ac)} -\: abc\;=\;60$
From (b), this is $11$

Therefore: . $6(11) - abc\;=\;60\quad\Rightarrow\quad \boxed{abc = 6}$

**Ruler of Hell** · June 29th 2006, 08:35 PM

Yay thanks! Yeah there was a typo... Phew. Um.... but um... I.... am sorry but I have another sum here... Err... I don't know what to say... :-(

The thing is a fraction, I couldn't type the code so I used the Underline tag to represent it as a fraction. I'm sorry..

Thats my last sum. If I post another single sum, slap me on the head. Sorry once again (especially to Soroban).

The sum is here:-
Find a³ + b³ + c³ - 3abc
......ab+bc+ac -a² - b² - c²

**topsquark** · June 30th 2006, 04:18 AM

Originally Posted by Ruler of Hell

Yay thanks! Yeah there was a typo... Phew. Um.... but um... I.... am sorry but I have another sum here... Err... I don't know what to say... :-(

The thing is a fraction, I couldn't type the code so I used the Underline tag to represent it as a fraction. I'm sorry..

Thats my last sum. If I post another single sum, slap me on the head. Sorry once again (especially to Soroban).

The sum is here:-
Find a³ + b³ + c³ - 3abc
......ab+bc+ac -a² - b² - c²

$\frac{a^3+b^3+c^3-3abc}{ab+bc+ac-a^2-b^2-c^2}$

Just click on the expression and a box will come up showing you how to code it.

-Dan

**Soroban** · June 30th 2006, 07:27 AM

Hello, Ruler of Hell!

That last one seems to just be an arithmetic problem . . .

Find . $\frac{a^3 + b^3 + c^3 - 3abc}{ab + bc + ac -a^2 - b^2 - c^2}$ .when $a = -5,\; b = -6,\;c=10$

We have: . $\frac{(-5)^3 + (-6)^3 + 10^3 - 3(-5)(-6)(10)}{(-5)(-6) + (-6)(10) + (-5)(10) - (-5)^2 - (-6)^2 - 10^2}$

. . . . . $= \;\frac{-125 - 216 + 1000 - 900}{30 - 60 - 50 - 25 - 36 - 100}$ $=\;\frac{-241}{-241}\;= \;\boxed{1}$

The numerator just happens to equal the denominator?

Well, there is a reason, but it depends on the values of $a,\,b,\,c$
. . . . . and my proof is truly awful!

**Ruler of Hell** · July 1st 2006, 01:26 AM

I got the exactly same thing, though I didn't directly substitute. First I cancelled the squares and cubes then I did the stuff. First I got zero, then I got a 1. I got afraid and so I left it out. Thanks Soroban! I owe you quite a lot! Yay, now to finish the final sum!!!! Hurrah! What joy!

Sorry if I wasted quite a lot of time of yours... and anybody else...

**Soroban** · July 1st 2006, 06:45 AM

Hello, Ruler of Hell!

I got the exactly same thing, though I didn't directly substitute.
First I cancelled the squares and cubes . . . No !!
Shame on you . . . you know that's illegal!

Okay, here's my clumsy proof . . .

Since $a = -5,\;b = -6,\;c = 10$ , we have: $a + b + c \:=\:-1$

. . Hence: . $a + b \:=\:-c -1$ [1]

Cube [1]: . $(a + b)^3\;=\;(-c-1)^3$

. . . $a^3$ $+ 3a^2b + 3ab^2 + b^3\;=\;-c^3 - 3c^2 - 3c - 1$

Arrange terms: . $a^3 + b^3 + c^3 + 3a^b + 3ab^2 \;= \;-3c^2 - 3c - 1$

Factor: . $a^3 + b^3 + c^3 + 3ab\underbrace{(a + b)} \;= \;-3c^2 - 3c - 1$

. . . . . $a^3 + b^2 + c^3 + 3ab(-c - 1) \;=\;-3c^2 - 3x - 1$

. . . . . . $a^3 + b^3 + c^3 - 3abc - 3ab \;= \;-3c^2 - 3c - 1$

. . . . . . . . . . . $a^3 + b^3 + c^3 - 3ab \;= \;3ab - 3c^2 - 3c - 1$

Hence: . $\boxed{Numerator \;= \;3ab - 3c^2 - 3c - 1}$

Square [1]: . $(-c-1)^2\;=\;(a + b)^2$

. . . . . . . . $c^2 + 2c + 1 \;= \;a^2 + 2ab + b^2$

. . . . . . . . . $-a^2 - b^2 \;= \; 2ab - c^2 - 2c - 1$

Add $ab + bc + ac - c^2$ to both sides:

. . . $(ab + bc + ac - c^2) - a^2 - b^2 \;= \;(ab + bc + ac - c^2)$ $+ 2ab - c^2 - 2c - 1$

. . . . $ab + bc + ac - a^2 - b^2 - c^2 \;= \;3ab + ac + bc - 2c^2 - 2c - 1$

We have: . $Denominator \;= \;3ab + c\underbrace{(a + b)} - 2c^2 - 2c - 1$

. . . . . . . . . . . . . . . . . . . $= \;3ab + c(-c - 1) - 2c^2 - 2c - 1$

. . . . . . . . . . . . . . . . . . . $= \;3ab - c^2 - c - 2c^2 - 2c - 1$

Hence: . $\boxed{Denominator \;= \;3ab - 3c^2 - 3c - 1}\quad\hdots\quad QED\,!$
.

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