यदि \(\rm k^4+\frac{1}{k^4}=194\) है, तो \(\rm k^3+\frac{1}{k^3}\) का मान ज्ञात कीजिए।

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Question

This question was previously asked in

SSC CGL 2022 Tier-I Official Paper (Held On : 02 Dec 2022 Shift 3)

Answer 1

दिया गया है:

\(\rm k^4+\frac{1}{k^4}=194\)

प्रयुक्त सूत्र:

\(\rm (a+\frac{1}{a})^2=\rm a^2+\frac{1}{a^2} + 2\)

यदि

\(\rm (a+\frac{1}{a})=b\)

तो,

\(\rm a^3+\frac{1}{a^3} =b^3-3b\)

गणना:

\(\rm k^4+\frac{1}{k^4}=194\)

⇒ \(\rm k^4+\frac{1}{k^4}+2=194+2\)

⇒ \(\rm k^4+\frac{1}{k^4}+2=196\)

⇒ \(\rm( k^2+\frac{1}{k^2})^2=14^2\)

⇒ \(\rm k^2+\frac{1}{k^2}=14\)

⇒ \(\rm k^2+\frac{1}{k^2}+2=14+2\)

⇒ \(\rm (k+\frac{1}{k})^2=16\)

⇒ \(\rm (k+\frac{1}{k})^2=4^2\)

⇒ \(\rm k+\frac{1}{k}=4\)

अब,

\(\rm k^3+\frac{1}{k^3}\) = 4³ - 3 × 4

⇒ \(\rm k^3+\frac{1}{k^3}\) = 64 - 12

⇒ \(\rm k^3+\frac{1}{k^3}\) = 52

∴ अभीष्ट उत्तर 52 है।

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