Question

evaluate , please give correct answer
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Answer 1

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$\sqrt[3]{288 \sqrt[3]{72 \sqrt[3]{27} } }$

$\sqrt[3]{288 \sqrt[3]{72 \sqrt[3]{ {3}^{3} } } }$

$\sqrt[3]{288 \sqrt[3]{72 \times 3 } }$

$\sqrt[3]{288 \sqrt[3]{216} }$

$\sqrt[3]{288 \sqrt[3]{ {6}^{3} } }$

$\sqrt[3]{288 \times 6}$

$\sqrt[3]{1728}$

$\sqrt[3]{ {12}^{3} }$

$12$

Answer by *BIG BRIAN*

Answer 2

Answer:

The value is 12

Step-by-step explanation:

Given

$\sqrt[3]{288 \sqrt[3]{72 \sqrt[3]{27} } }$

we know the cube of 3 is 27 so the last cube root cancels out and the it becomes

$\sqrt[3]{288 \sqrt[3]{72 \times 3} }$

Now 72 x 3 equals 216 which is the cube of 6 hence the 2nd cube root cancels out and the answer comes out to be

$\sqrt[3]{288 \times 6}$

Now on factorizing 288 we get it as 2 x 2 x 2 x 2 X 2 x 3 x 3 and the last 6 equals 2 x 3

So, we get

$\sqrt[3]{ {2}^{6} \times {3}^{3} }$

On removing the cube root we get the value

2² x 3 = 12