<s>[INST] If $a + b = \frac{1}{4}$, what is $a^3 + b^3$? I just want to know how to solve this problem. 
If $a + b = \dfrac{1}{4}$, what is $a^3 + b^3$ equals to?

My work - but still doesn't give me an answer. 
$a^2 + b^2 = (a + b)^2 - 2ab = \dfrac{1}{16} - 2ab$ 
$a^3 + b^3 = (a + b)^3 - 3ab(a + b) = \dfrac{1}{64} - \dfrac{3ab}{4}$.
How can I get the value for $ab$ here?
 [/INST] The value of $y = a^3+b^3$ contains a family of solutions (like a curve). You cannot establish a value for $y$ unless something else is specified.
For example besides $a+b = \tfrac{1}{4}$, let us use some artbitrary value $a-b = \lambda$. The solve for $a$ and $b$ in terms of $\lambda$ and put them into $y$
$$ y = \left( \frac{1+4 \lambda}{8} \right)^3 + \left( \frac{1-4 \lambda}{8} \right)^3=\frac{1}{256} + \frac{3 \lambda^2}{16} = \frac{1}{256} + \frac{3 (a-b)^2}{16}$$
so the minimum value of y exists when a=b, and thus λ=0.
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