If 4x4 - 37x2 + 9 = 0, x > $\sqrt{\frac{3}{2}}$, then what is the value of 8x3 - $\frac{27}{x^3}$ ?
Answer & explanation
Correct answer: option 2
4x4 - 37x2 + 9 = 0
Let us consider x2 = m
= 4m2 – 37m + 9 = 0
= 4m2 – 36m - m + 9 = 0
= 4m(m - 9) - 1(m - 9) = 0
= (4m - 1)(m - 9) = 0
a = \(\frac{1}{4}\) or a = 9
a = x2 = 9
x = 3
8x3 - $\frac{27}{x^3}$ = 8 × 33 - $\frac{27}{3^3}$
= 216 - 1 = 215