Question

factorise 1/64a^2-4/3ab+256/9b^2​

Answer 1

Step-by-step explanation:

1/64a^2-4/3ab+256/9b^2

=(1/8a)^2-2×1/8a×16/3b+(16/3b)^2. (a^2-2ab+b^2=(a-b)^2)

=(1/8a-16/3b)^2

Answer 2

$\huge{\boxed{\mathcal\pink{\fcolorbox{red}{purple}{Solution:-}}}}$

Question:-

$\boxed{\frac{1}{64} - \frac{4}{3ab} + \frac{256}{9 {b}^{2}} }$

Answer:-

Factorising-

$= > \sqrt{ \frac{1}{64a {}^{2} } -\frac{265}{9b {}^{2} } } \\ \\ = > \frac{1}{8a {}^{2} } - \frac{16}{3b {}^{2} } \\ \\ = > ( { \frac{1}{8a} - \frac{16}{3b} })^{2}$

$\huge{\boxed{\mathcal\red{\fcolorbox{pink}{blue}{THANKS:-}}}}$