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