Question

if a+b=4 then, a^3+b^3=?

Answer 1

a+b=4 Make this one... a = -b + 4
a^3+b^3
Substitute the first equation into the second one
(-b + 4)^3 + b^3
Does this help?

Answer 2

Having this (-b + 4)^3 + b^3 you can now solve for b.

Answer 3

and then slove for a once u get b right?

Answer 4

Yepp by plugging b back into the first equation. (a+b=4)

Answer 5

oh okay thanks :D

Answer 6

"Having this (-b + 4)^3 + b^3 you can now solve for b."

But we don't know what a³ + b³ equal to..?

Answer 7

@geerky42 You have a point here...

Answer 8

But it also looks like an if-then statement. Now that I look at it,

Answer 9

Well when I do
(-b + 4)^3 + b^3
-b^3+4^3+b^3
b cancels out and it just sits there with 4^3
which is 64 so my way is for sure not the way to do it, if you try a this same way a will cancel out also.

Answer 10

But (-b + 4)³ \(\neq\) -b³ + 4³, it's actually -b³ + 12b² - 48b + 64

Answer 11

You can't do it...
Actually it depends on what a and b equal.

Mathematically:
(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3
= (a^3 + b^3) + 3ab(a+b)

therefore by rearranging

a^3 + b^3 = 3ab(a+b) - (a+b)^3

you have (a+b), but you are missing ab (which would mean you could calculate a and b)

If you don't get what I'm talking about, try an example:

Let a = 1, b = 3
a + b = 4, a^3+b^3 = 28

Now, Let a = 0, b = 4
a + b = 4, a^3 + b^3 = 64...a different answer.

So you can see that a^3 + b^3 depends entirely on what a and b are.
You don't have enough information to calculate this