Question

??
345+124
5/512
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$find \: the \: value \: of \: 3 \sqrt{512 } + 3 \sqrt{343} + 3 \sqrt{729 }$
​

Answer 1

$8 + 7 +9 = 24$

Cube root of 512 = 8

Cube root of 343 = 7

Cube root of 729 = 9

Hope it helped you : )

Answer 2

ANSWER

48$\sqrt{2}$ + 21$\sqrt{7}$ + 81

Step-by-step explanation:

3$\sqrt{512}$ + 3$\sqrt{343}$ + 3$\sqrt{729}$

Factor 512  = 16^2 * 2. Rewrite the square root of the product $\sqrt{16^2 * 2}$ as the product of square roots $\sqrt{16^2 } \sqrt{2\\}$. Take the square root of 16^2.

3 *16$\sqrt{2}$ + 3$\sqrt{343}$ + 3$\sqrt{729}$

Multiply 3 and 16 to get 48

48$\sqrt{2}$ + 3$\sqrt{343}$ + 3$\sqrt{729}$

Factor 343 = 7^2 * 7. Rewrite the  square root of the product $\sqrt{7^2 * 7}$ as the product of square roots  $\sqrt{7^2}$$\sqrt{7}$. Take the square root of 7^2.

48$\sqrt{2}$ + 21 + 3 * 7$\sqrt{7}$ + 3$\sqrt{729}$

Multiply 3 and 7 to get 21.

Calculate the square root of 729 and get 27.

42$\sqrt{2}$ + 21$\sqrt{7\\}$ + 3* 27

Multiply 3 and 27  to get 81.

48$\sqrt{2}$ + 21$\sqrt{7}$ + 81