Question of The Day17-05-2021

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 = a2 + b2 + 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 = a3 + b3 + 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

⇒ a3 = 12 +3a

⇒ a3 – 3a = 18

⇒ a (a2 - 3) = 18

⇒ a = 3

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

Hence, (a) is the correct answer.

Such type of question is asked in various government exams like SSC CGL, SSC MTS, SSC CPO, SSC CHSL, RRB JE, RRB NTPC, RRB GROUP D, RRB OFFICER SCALE-I, IBPS PO, IBPS SO, RRB Office Assistant, IBPS Clerk, RBI Assistant, IBPS RRB OFFICER SCALE 2&3, UPSC CDS, UPSC NDA etc.

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