Question

find the square root of the sum of the 1st 3 natural no.​

Answer 1

Step-by-step explanation:

$sum of cubes</p><p></p><p>1 {}^{3} + 2 {}^{3} + 3 {}^{3} 1 </p><p>3</p><p> +2 </p><p>3</p><p> +3 </p><p>3</p><p> </p><p></p><p>1 + 8 + 271+8+27</p><p></p><p> \\ 3636 \\ </p><p></p><p>Square \: root \: of \: sum \: of \: first \: 3 \: natural \: numbers</p><p></p><p> \\ \sqrt{36} </p><p> \\ 36</p><p> </p><p> </p><p></p><p>= 6=6$