+0  
 
0
33
1
avatar

Let a and b be real numbers such that a^3 + 3ab^2 = 679 and a^3 - 3ab^2 = 673.  Find a - b.

 Jan 6, 2023
 #1
avatar+115 
0

Alright, the first step is very simple

So basically subtract 6ab^2 from the first equation and you will get the second equation so 6ab^2 = 6

\(6ab^2 = 6\)

\(ab^2 = 1\)

now, you are asking for a - b

\(b^2 = 1/a\)

\(\cdot 1/b\) on both sides

\(b = 1/ab\)

im just doing every possible thing i can do bc this is my first time seeing this so im doing it as well as you sorry

\(ab = 1/b\) ok that might be useful

oh......

\(a^3 + 3 = 679\)

\(a^3 = 676\)

\(a \approx 8.77639\)

so b is 1/ whatever that is. Square rooted.....

its approximately 8.43883.... probably not correct, looking at the numbers. Maybe its like the square root of something?

 its like \(\sqrt{71.212}\) not exacly but close

 Jan 6, 2023

18 Online Users

avatar
avatar