resolve into factor

**rizwmun** · February 13th 2013, 10:36 AM

1. 2a²b²+2b²c²+2c²a²-a⁴-b⁴-c⁴2. a³+1/a³=18*3^1/2 ,prove that, x = 3^1/2+ 2^1/2

**Soroban** · February 13th 2013, 12:44 PM

Hello, rizwmun!

$\text{1. Factor: }\:2a^2b^2 + 2b^2c^2 + 2a^2c^2 - a^4 - b^4 - c^4$

$(a+b+c)(a+b-c)(a-b+c)(-a+b+c)$

$\text{Given: }\:x^3 + \frac{1}{x^3} \:=\:18\sqrt{3},\;\text{prove that: }\:x \:=\:\sqrt{3} + \sqrt{2}$

$x^3 \;=\;(\sqrt{3}+\sqrt{2})^3 \;=\;3\sqrt{3} +3(3\!\cdot\!\sqrt{2}) + 3(\sqrt{3}\!\cdot\!2) + 2\sqrt{2} \;=\;9\sqrt{3}+11\sqrt{2}$

$\frac{1}{x^3} \;=\;\frac{1}{9\sqrt{3}+11\sqrt{2}}\cdot {\color{blue}\frac{9 \sqrt{3}-11\sqrt{2}}{9\sqrt{3}-11\sqrt{2}}} \;=\; \frac{9\sqrt{3}-11\sqrt{2}}{243-242} \;=\;9\sqrt{3} - 11\sqrt{2}$

Therefore: . $x^3 + \frac{1}{x^3} \;=\;\left(9\sqrt{3} + 11\sqrt{2}\right) + \left(9\sqrt{3} - 11\sqrt{2}\right) \;=\;18\sqrt{3}$

**rizwmun** · February 14th 2013, 10:23 AM

thanks for replay.....but for nunmer 2, I have to prove the last portion from first portion not by last portion.
And I need whole procedure for number 1.

**Soroban** · February 14th 2013, 03:56 PM

Hello agaibn, rizwmun!

I have a method for #2 ... which is a bit devious.
Perhaps someone else can find an elegant solution.

$\text{2. Given: }\,x^3+\frac{1}{x^3} \:=\:18\sqrt{3}\;\;{\color{blue}[1]}$

. . . $\text{ Prove: }\,x \:=\:\sqrt{3}+\sqrt{2}$

Let $x + \frac{1}{x} \:=\:a\;\;{\color{blue}[2]}$

Cube both sides: . $\left(x + \frac{1}{x}\right)^3 \:=\:a^3 \quad\Rightarrow\quad x^3 + 3x + \frac{3}{x} + \frac{1}{x^3} \:=\:a^3$

. . . . . $\left(x^3 + \frac{1}{x^3}\right) + 3\underbrace{\left(x + \frac{1}{x}\right)}_{\text{This is }a} \:=\:a^3 \quad\Rightarrow\quad x^3 + \frac{1}{x^3} + 3a \:=\:a^3$

. . . . . $x^3+\frac{1}{x^3} \:=\:a^3 - 3a$

From [1], we have: . $a^3 - 3a \:=\:18\sqrt{3} \quad\Rightarrow\quad a^3 - 3a - 18\sqrt{3} \:=\:0$

We find that the only real root is:. $a \:=\:2\sqrt{3}$

Substitute into [2]: . $x + \frac{1}{x} \:=\:2\sqrt{3} \quad\Rightarrow\quad x^2 - 2\sqrt{3}x + 1 \:=\:0$

Quadratic Formula: . $x \;=\;\frac{2\sqrt{3} \pm\sqrt{(2\sqrt{3})^2 - 4(1)(1)}}{2(1)}$

. . . . . . . . . . . . . . . $x\;=\;\frac{2\sqrt{3} \pm\sqrt{8}}{2} \;=\;\frac{2\sqrt{3}\pm2\sqrt{2}}{2}$

Therefore: . $x \:=\:\sqrt{3} \pm\sqrt{2}$

**RolandoGamble** · February 15th 2013, 09:38 PM

thanks for replay.....but for nunmer 2, I have to prove the last portion from first portion not by last portion.

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