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Mathematics 14 Online
OpenStudy (anonymous):

if a+b=4 then a3+b3=?

OpenStudy (aravindg):

is that the only data given ??

Parth (parthkohli):

Go go go Govinda!

Parth (parthkohli):

Do you know what \(ab\) is?

OpenStudy (agent0smith):

There's many solutions...

Parth (parthkohli):

There's infinite solutions in fact...

OpenStudy (agent0smith):

Yeah actually there would be, I think even if it's limited to integers

OpenStudy (anonymous):

Do you have any extra info? Right now there are infinitely many solutions as Parthkohli already mentioned

OpenStudy (anonymous):

No extra information is there Mr. Meepi.

OpenStudy (anonymous):

a+b=4 .. a^3+b^3=4^3? So a^3+b^3=64

OpenStudy (anonymous):

Doesn't work that way: a + b = 4 gives (a + b)^3 = 4^3 \[\left(a+b\right)^3=\left(\begin{matrix}3 \\ 0\end{matrix}\right)a^3+\left(\begin{matrix}3 \\ 1\end{matrix}\right)a^2b + \left(\begin{matrix}3 \\ 2\end{matrix}\right)ab^2+\left(\begin{matrix}3 \\ 3\end{matrix}\right)b^3=a^3+3a^2b+3ab^2+b^3\] IE: \[a^3 + b^3=64-3ab(a+b)=64-12ab\] But you can't get any further than that since you don't know the product ab

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