If\(\sqrt x+{1 \over \sqrt x}=20\), then find the value of \(\sqrt[3]{x}+{1 \over \sqrt[3]{x}}\)

## Answer

Correct Answer : a ) 3

Explanation :According to the question

\(\sqrt x+{1 \over \sqrt x}=20\)

We know that

(a + b)^{2} = a^{2} + b^{2} + 2ab

\(\sqrt x+{1 \over \sqrt x}=20\)

Squaring on both side

⇒\((\sqrt x+{1 \over \sqrt x})^2=(\sqrt 20) ^2\)

⇒\(x+{1 \over x}+2×{1 \over x}×x=20\)

⇒\(x+{1 \over x}=20-2\)

⇒\(x+{1 \over x}=18..................(1)\)

We know that

(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)

Let \(\sqrt[3]{x}+{1 \over \sqrt[3]{x}}=a\)

\(\sqrt[3]{x}+{1 \over \sqrt[3]{x}}\)

On cubing the above equation

⇒\((\sqrt[3]{x} +{1 \over \sqrt[3]{x}})^3=(\sqrt[3]{x})^3+({1 \over \sqrt[3]{x}})^3+3*\sqrt[3]{x}*{1 \over \sqrt[3]{x}}*(\sqrt[3]{x}+{1 \over \sqrt[3]{x}})\)

⇒ \((\sqrt[3]{x} +{1 \over \sqrt[3]{x}})^3={x}+{1 \over{x}}+3(\sqrt[3]{x}*{1 \over \sqrt[3]{x}})\)

Substituting the values

⇒ a^{3} = 12 +3a

⇒ a^{3} – 3a = 18

⇒ a (a^{2} - 3) = 18

⇒ a = 3

⇒\(\sqrt[3]{x}+{1 \over \sqrt[3]{x}}=3\)

Hence, (a) is the correct answer.

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