Question

[1/4b+1/3c] square expand using identities
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Answer 1

Answer:

$( \frac{1}{4b} + \frac{1}{3c}) {}^{2} \\ \\ = { (\frac{1}{4b}) }^{2} + ( \frac{1}{3c} {)}^{2} + 2 \times \frac{1}{4b} \times \frac{1}{3c} \\ \\ = \frac{1}{16 {b}^{2} } + \frac{1}{9 {c}^{2} } + \frac{2}{12bc} \\ \\ = \frac{1}{16 {b}^{2} } + \frac{1}{9 {c}^{2} } + \frac{1}{6bc}$

Answer 2

(1/4b+1/3c)²

= (1/4b)²+ (1/3c)² + 2*1/4b*1/3c

= 1/16b²+ 1/9c²+2/12bc

Step-by-step explanation:

Hope it helps you have a great day :-))