Question

Show that.
Third one....
(With method)

Answer 1

Using formula of (a+b)^2

(16/3)m^2+(9/16)n^2-(2mn)+2mn [as (4/3)*(3/4)=1]

now

2mn-2mn=0

therefore ans

=(16/3)m^2+(9/16)n^2

MARK BRAINLIEST

FOLLOW.

Answer 2

Your answer

$\frac{4}{3}m( \frac{4}{3}m - \frac{3}{4}n) - \frac{3}{4}n( \frac{4}{3}m - \frac{3}{4} n) + 2mn = \frac{16}{9} {m}^{2} + \frac{9}{16} {n}^{2} \\ \\ \frac{16}{9} {m}^{2} - \frac{12}{12} mn - \frac{12}{12} mn + \frac{9}{16} {n}^{2} + 2mn = \frac{16}{9} {m}^{2} + \frac{9}{16} {n}^{2} \\ \\ \frac{16}{9} {m}^{2} - 2mn + \frac{9}{16} {n}^{2} + 2mn = \frac{16}{9} {m}^{2} + \frac{9}{16} {n}^{2} \\ \\ \frac{16}{9} {m}^{2} - 2mn + 2mn + \frac{9}{16} {n}^{2} = \frac{16}{9} {m}^{2} + \frac{9}{16} {n}^{2} \\ \\ therefore \: your \: answer \: is \: \\ \frac{16}{9} {m}^{2} + \frac{9}{16} {n}^{2} = \frac{16}{9} {m}^{2} + \frac{9}{16} {n}^{2}$