Question

(√12a+√6b) whole square​

Answer 1

Step-by-step explanation:

$( \sqrt{12a } + \sqrt{6b} ) {}^{2} \\ 12a + 6b + \sqrt{24ab}$

I hope it helps you

Answer 2

Answer:

$( \sqrt{12a} + \sqrt{6b} ) {}^{2} \\ {( \sqrt{a} + \sqrt{b} ) }^{2} = {( \sqrt{a} )}^{2} + 2 \sqrt{ab} \: + { (\sqrt{b} )}^{2} \\ = {(\sqrt{12a})}^{2} + (2 \times \sqrt{12a} \times\sqrt{6b}) + {(\sqrt{6b})}^{2} \\ = 12a + 2 \sqrt{72ab} + 6b \\ = 12a + 2 \sqrt{2 \times 2 \times 2 \times 3 \times 3 \times a \times b} + 6b \\ = 12a + 2 \sqrt{ {2}^{2} \times 2 \times {3}^{2} \times a \times b } + 6b \\ = 12a + 2 \sqrt{ {6}^{2} \times 2 \times a \times b} + 6b \\ = 12a +( 2 \times 6 \sqrt{2 \times a \times b} ) + 6b \\ = 12a + 12 \sqrt{2ab} + 6b$