PROVE THAT:
$\displaystyle 12^3$=(9+$\displaystyle sqrt{5}$)^3+(9-$\displaystyle SQRT{5}$)^3
2. $\displaystyle e^ln36$=36 why?
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, $\displaystyle x = \ln y \iff y = e^x$, if $\displaystyle y > 0$ (it's undefined otherwise).
So we have $\displaystyle e^x = e^{\ln y}$ (doing the same thing to both sides of the first of the above).
Then we note that from the second of the above, $\displaystyle y = e^x$ and so it follows directly that:
$\displaystyle y = e^{\ln y}$
That holds for all $\displaystyle y > 0$ and so definitely holds for $\displaystyle y = 36$.
Did you try just doing the arithmetic on the right?
$\displaystyle (a+b)^3= a^3+ 3a^2b+ 3ab^2+ b^3$
So $\displaystyle (9+ \sqrt{5})^3= 9^3+ 3(9^2)(\sqrt{5})+ 3(9)(\sqrt{5})^2+ (\sqrt{5})^3$
$\displaystyle = 729+ 243\sqrt{5}+ 135+ 5\sqrt{5}= 964+ 248\sqrt{5}$
$\displaystyle (9- \sqrt{5})^3= 9^3- 3(9^2)(\sqrt{5})+ 3(9)(\sqrt{5})^2- (\sqrt{5})^3$
$\displaystyle = 729- 243\sqrt{5}+ 135- 5\sqrt{5}= 964- 248\sqrt{5}$
so the terms involving $\displaystyle \sqrt{5}$ cancel and the sum is 1728.
And, of course, $\displaystyle 12^3= 1728$
You could also have seen this by observing that the odd powers of $\displaystyle \sqrt{5}$ will cancel while the even powers add:
$\displaystyle 2(9^3+ 3(9)(5))= 2(9)(9^2+ 15)= 2(9)(81+5)$[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