Prove: \[ \left(\frac{1}{4}\right)^{-2} – 3\times 8^{2/3} \times 4^0 + \left(\frac{9}{16}\right)^{-1/2} = \frac{16}{3} \]
Solution
\[ \left(\frac{1}{4}\right)^{-2} – 3\times 8^{2/3} \times 4^0 + \left(\frac{9}{16}\right)^{-1/2} \]
\[ = 4^2 – 3\times (2^3)^{2/3} \times 1 + \left(\frac{3^2}{4^2}\right)^{-1/2} \]
\[ = 16 – 3\times 2^2 + \left(\frac{3}{4}\right)^{-1} \]
\[ = 16 – 12 + \frac{4}{3} \]
\[ = 4 + \frac{4}{3} \]
\[ = \frac{16}{3} \]