if a+b=4 then a3+b3=?
is that the only data given ??
Go go go Govinda!
Do you know what \(ab\) is?
There's many solutions...
There's infinite solutions in fact...
Yeah actually there would be, I think even if it's limited to integers
Do you have any extra info? Right now there are infinitely many solutions as Parthkohli already mentioned
No extra information is there Mr. Meepi.
a+b=4 .. a^3+b^3=4^3? So a^3+b^3=64
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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