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Bunuel
What is the value of \((\frac{1}{\sqrt{3}})^3\)?

A. \(\frac{\sqrt{3}}{9}\)

B. \(\frac{3}{\sqrt{3}}\)

C. 1

D. \(\sqrt{3}\)

E. 3

\((\frac{1}{\sqrt{3}})^3\)

= \(\frac{√3^3}{3^3}\)

= \(\frac{3√3}{27}\)

= \(\frac{√3}{9}\)

Thus, answer will be (A)
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Bunuel
What is the value of \((\frac{1}{\sqrt{3}})^3\)?

A. \(\frac{\sqrt{3}}{9}\)

B. \(\frac{3}{\sqrt{3}}\)

C. 1

D. \(\sqrt{3}\)

E. 3
In steps. \((\frac{1}{\sqrt{3}})^3\) =

1. Multiply the term by itself once:

\((\frac{1}{\sqrt{3}})*(\frac{1}{\sqrt{3}})\) =

\((\frac{1*1}{\sqrt{3}*\sqrt{3}})\) =

\(\frac{1}{3}\)

2. Then multiply the product \(\frac{1}{3}\)by the original term

\((\frac{1}{3})\)* \((\frac{1}{\sqrt{3}})\) = \(\frac{1}{3\sqrt{3}}\).

Rationalize the denominator:

\((\frac{1}{3\sqrt{3}})\) * \((\frac{\sqrt{3}}{\sqrt{3}})\)

= \((\frac{1*\sqrt{3}}{3\sqrt{3}*\sqrt{3}})\)

= \(\frac{\sqrt{3}}{3*3}\)

= \(\frac{\sqrt{3}}{9}\)
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Bunuel
What is the value of \((\frac{1}{\sqrt{3}})^3\)?

A. \(\frac{\sqrt{3}}{9}\)

B. \(\frac{3}{\sqrt{3}}\)

C. 1

D. \(\sqrt{3}\)

E. 3

We need to rationalize the denominator in the given expression.

Since (√3)^3 = (√3)^2 x √3 = 3√3, (1/√3)^3 = 1/3√3.

Multiplying this fraction by √3/√3, we have:

1/3√3 x √3/√3 = √3/(3 x 3) = √3/9

Answer: A
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\((\frac{1}{\sqrt{3}})^3\)

Lets simplify this, first step is cube the numerator and denominator.

\({1^3 /{\sqrt{3} * \sqrt{3} * \sqrt{3}}\)

\({1/3 * \sqrt{3}}\)

Multiply Numerator and Denominator by \(\sqrt{3}\)


\({1 * \sqrt{3}/3 * \sqrt{3} * \sqrt{3}}\)

\(\frac{\sqrt{3}}{9}\)

Hence, Answer is A
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\((\frac{1}{\sqrt{3}})^3\) = \(\frac{1}{\sqrt{3}}*\frac{1}{\sqrt{3}}*\frac{1}{\sqrt{3}}\) = \(\frac{1}{(3*\sqrt{3})}\)

Multiplying and dividing by \(\sqrt{3}\)
= \(\frac{1}{{3*\sqrt{3}*\sqrt{3}}}*\sqrt{3}\) = \(\frac{\sqrt{3}}{9}\) (Option A)
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