If (4a - 3b) = 1, $ab = \frac{1}{2}$, where a > 0 and b > 0, what is the value of $(64a^3 + 27b^3)$?

35

If x + \(\frac{1}{x}\)  = n

then, $x^3 +\frac{1}{x^3}$ = n³ - 3 × n

If x - \(\frac{1}{x}\)  = n

then 

then, x + \(\frac{1}{x}\)  = \(\sqrt {n^2 + 4}\)

If (4a - 3b) = 1,

$ab = \frac{1}{2}$

what is the value of $(64a^3 + 27b^3)$

(4a + 3b) = \(\sqrt {1^2 + 2 \times 4 \times 3}\) = 5

The value of $(64a^3 + 27b^3)$ = 5³ - 3 × 5 × 4 × 3  × $\frac{1}{2}$  = 125 - 90 = 35