# Algebra

• Nov 7th 2006, 03:38 PM
Rimas
Algebra
If:
X+1/X=4

then what is the value of

X^3+1/x^3
• Nov 7th 2006, 03:44 PM
Dergyll
Hi

Im just going to do this the classic/stupid way.

Solve for x in x+1/x=4 and you get x=1/3, then substitude it into the other equation and bam you got the answer.

THere's probably a better, more efficient way though...

Derg
• Nov 7th 2006, 04:23 PM
Soroban
Hello, Rimas!

This problem has a spectacular solution . . .

Quote:

If $\displaystyle x + \frac{1}{x} \,=\,4$

then what is the value of $\displaystyle x^3 + \frac{1}{x^3}$

Cube the equation: .$\displaystyle \left(x + \frac{1}{x}\right)^3\;=\;4^3$

We have: .$\displaystyle x^3 + 3x^2\!\cdot\!\frac{1}{x} + 3x\!\cdot\!\frac{1}{x^2} + \frac{1}{x^3} \:= \:64\quad\Rightarrow\quad x^3 + 3x + \frac{3}{x} + \frac{1}{x^3}\:=\:64$

Then: .$\displaystyle x^3 + 3\underbrace{\left(x + \frac{1}{x}\right)}_\downarrow + \frac{1}{x^3} \:=\:64$
. . . . . . . $\displaystyle x^2 + 3(4) + \frac{1}{x^3} \:=\:64$

. . . . . . . . . $\displaystyle x^3 + \frac{1}{x^3} \:=\:52$