यदि \(x=\sqrt{6}+\sqrt{5}\), तो \(x^3+\frac{1}{x^3}\) का मान और प्रकृति क्या है?
If \(x=\sqrt{6}+\sqrt{5}\), what is the value and nature of \(x^3+\frac{1}{x^3}\)?
Explanation opens after your attempt
A. \(42\sqrt{6}\), अपरिमेय\(42\sqrt{6}\), irrational
Concept
(\(\sqrt{6}+\sqrt{5}\)\(\sqrt{6}-\sqrt{5}\)=1), so \(\frac{1}{x}=\sqrt{6}-\sqrt{5}\).
Why this answer is correct
\(x+\frac{1}{x}=2\sqrt{6}\), hence (x-3+\frac{1}{x-3}=\(2\sqrt{6}\)3-3\(2\sqrt{6}\)=42\sqrt{6}).
Exam Tip
In cube-type questions, finding \(x+\frac{1}{x}\) first is the easier method. चरण 1: (\(\sqrt{6}+\sqrt{5}\)\(\sqrt{6}-\sqrt{5}\)=1), इसलिए \(\frac{1}{x}=\sqrt{6}-\sqrt{5}\)। चरण 2: \(x+\frac{1}{x}=2\sqrt{6}\), अतः (x-3+\frac{1}{x-3}=\(2\sqrt{6}\)3-3\(2\sqrt{6}\)=42\sqrt{6})। चरण 3: घन वाले प्रश्नों में पहले \(x+\frac{1}{x}\) निकालना आसान तरीका है।
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