Help with an algegra equation

[UncleTodd]

I'm trying to help my nephew with his algebra homework, but it has been way too many years since I had algebra. We have a problem that has me stumped. I'm not sure if it qualifies as beginning or intermediate. At first glance, it looks easy, but I must be missing something obvious, because I can't figure it out!

Anyway, if anybody could help figure this out (and explain it), it would be very much appreciated.

Here is the equation:

a + b = 3
a squared + b squared = 6
what is the value of a to the third power + b to the third power?


Thanks for any help,

Todd

 

[wjm11]

a + b = 3
a squared + b squared = 6
what is the value of a to the third power + b to the third power?

a + b = 3
a^2 + b^2 = 6

Solve for “a” in the first equation:

a = 3 – b

Substitute into the 2nd equation:

(3 – b)^2 + b^2 = 6

Expand, simplify:

9 – 6b + b^2 + b^2 = 6
2b^2 – 6b + 9 = 6
2b^2 –6b +3 = 0

And solve for b, using the quadratic formula:

b = [-(-6) +/- (36 – 4(2)(3))^.5]/(2(2))
b = [6 +/- (12)^.5] / 4
b = [3 +/- (3)^.5]/2

Substitute the solutions back into the first equation to find “a”.

Therefore, when b = [3 + (3)^.5]/2, a = [3 - (3)^.5]/2

Now that you have a and b, you can cube them for your final result:

([3 + (3)^.5]/2)^3 + ([3 - (3)^.5]/2)^3 = 13.245 + .2548 = 13.5

Hope that helps.

 

[Deleted member 4993]

I'm trying to help my nephew with his algebra homework, but it has been way too many years since I had algebra. We have a problem that has me stumped. I'm not sure if it qualifies as beginning or intermediate. At first glance, it looks easy, but I must be missing something obvious, because I can't figure it out!

Anyway, if anybody could help figure this out (and explain it), it would be very much appreciated.

Here is the equation:

a + b = 3
a squared + b squared = 6
what is the value of a to the third power + b to the third power?


Thanks for any help,

Todd

There are many ways to solve a problem. Bill showed you the direct way to solve the problem - it will work everytime.
But for this problem - there is a short-cut (sort of).

(a +b)[sup:1o0h8wn7]2[/sup:1o0h8wn7] = a[sup:1o0h8wn7]2[/sup:1o0h8wn7] + b[sup:1o0h8wn7]2[/sup:1o0h8wn7] + 2ab

9 = 6 + 2ab

ab = 3/2

a[sup:1o0h8wn7]3[/sup:1o0h8wn7] + b[sup:1o0h8wn7]3[/sup:1o0h8wn7] = (a+b)(a[sup:1o0h8wn7]2[/sup:1o0h8wn7] + b[sup:1o0h8wn7]2[/sup:1o0h8wn7] - ab) = 3(6-3/2) = 3*4.5 = 13.5

In timed tests (like SAT, GRE, etc.) - these things can come handy.

 

[soroban]

Hello, UncleTodd!

Subhotosh is absolutely correct.
Here's my approach to that shortcut.

\(\displaystyle \begin{array}{cccc}a + b &=& 3 & [1] \\ a^2 + b^2 &=& 6 & [2]\end{array}\)

\(\displaystyle \text{Find the value of: }\: a^3+ b^3\)

\(\displaystyle \text{Square [1]: }\;(a+b)^2 \:=\:3^2 \quad\Rightarriw\quad a^2 + 2ab + b^2 \:=\:9\)


\(\displaystyle \text{We have: }\;\underbrace{a^2+b^2}_{\text{This is 6}} + 2ab \:=\:9 \quad\Rightarrow\quad 2ab \:=\:3 \quad\Rightarrow\qud ab \:=\:\tfrac{3}{2}\)


\(\displaystyle \text{Cube [1]: }\;(a+b)^3 \:=\:3^3 \quad\Rightarrow\quad a^3 + 3a^2b + 3ab^2 + b^3 \:=\:27\)


\(\displaystyle \text{We have: }\;a^3+b^3 + 3\underbrace{(ab)}_{\frac{3}{2}}\underbrace{(a+b)}_{3} \:=\:27 \quad\Rightarrow\quad a^3+b^3 + \tfrac{27}{2} \:=\:27\)


\(\displaystyle \text{Therefore: }\;a^3 + b^3 \;=\;\frac{27}{2}\)

 

[UncleTodd]

Thanks a ton guys! This was exactly what I was looking for. Especially the break down of HOW to figure it out. It helped bring back my algebra from years ago enough that I think I can explain it to my nephew now!

Thanks again, you guys rock!

Todd