Question

9 - root 3 bracket cube=?

Answer 1

Step-by-step explanation:

It is given that,

$9 - ({ \sqrt{3} })^{3}$

By expanding ,

We get,

$3 \times 3 - 3 \times \sqrt{3}$

Taking 3 as common factor,

$3(3 - \sqrt{3} )$

Again Expand ,

$3( \sqrt{3 } \times \sqrt{3} - \sqrt{3} )$

Taking Common again,

$3 \sqrt{3} ( \sqrt{3} - 1)$

Hence this is Answer,

If Cube is of Whole,

i.e.,

$( {9 - \sqrt{3} })^{3}$

Then , By using Identity,

$({a - b})^{3} = {a}^{3} - {b}^{3} - 3 {a}^{2} b + 3a {b}^{2}$

Putting the values in this Formula,

${9}^{3} - ({ \sqrt{3} })^{3} - 3 ({9})^{2} \sqrt{3} + 3 \times 9 \times ({ \sqrt{3} })^{2}$

$729 - 3 \sqrt{3} - 3 \times 81 \times \sqrt{3} + 3 \times 9 \times 3$

$729 - 3 \sqrt{3} - 243 \sqrt{3} + 81$

$810 - 246 \sqrt{3}$

So this is Answer,

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