# Thread: 2 simple Algebra questions

1. ## 2 simple Algebra questions

PROVE THAT:
$12^3$=(9+ $sqrt{5}$)^3+(9- $SQRT{5}$)^3
2. $e^ln36$=36 why?

2. The first one's giving me a headache, so I'll waffle at the second.

By definition, the "taking the logarithm of" function and the "raising e to the power of" are direct inverses.

That is, by definition, $x = \ln y \iff y = e^x$, if $y > 0$ (it's undefined otherwise).

So we have $e^x = e^{\ln y}$ (doing the same thing to both sides of the first of the above).

Then we note that from the second of the above, $y = e^x$ and so it follows directly that:

$y = e^{\ln y}$

That holds for all $y > 0$ and so definitely holds for $y = 36$.

3. thanx!

4. Originally Posted by blertta
PROVE THAT:
$12^3$=(9+ $sqrt{5}$)^3+(9- $SQRT{5}$)^3
Did you try just doing the arithmetic on the right?

$(a+b)^3= a^3+ 3a^2b+ 3ab^2+ b^3$

So $(9+ \sqrt{5})^3= 9^3+ 3(9^2)(\sqrt{5})+ 3(9)(\sqrt{5})^2+ (\sqrt{5})^3$
$= 729+ 243\sqrt{5}+ 135+ 5\sqrt{5}= 964+ 248\sqrt{5}$

$(9- \sqrt{5})^3= 9^3- 3(9^2)(\sqrt{5})+ 3(9)(\sqrt{5})^2- (\sqrt{5})^3$
$= 729- 243\sqrt{5}+ 135- 5\sqrt{5}= 964- 248\sqrt{5}$
so the terms involving $\sqrt{5}$ cancel and the sum is 1728.

And, of course, $12^3= 1728$

You could also have seen this by observing that the odd powers of $\sqrt{5}$ will cancel while the even powers add:
$2(9^3+ 3(9)(5))= 2(9)(9^2+ 15)= 2(9)(81+5)$[tex]= 2(9)(96)= 2(3^2)(4)(24)= 2(3^2)(4)(4)(2)(3)= (2^6)(3^3)= (4^3)(3^3)= ((4)(3))^2= (12)^3