KnowledgeBoat Logo
|

Mathematics

If x2+1x=313\dfrac{x^2 + 1}{x} = 3\dfrac{1}{3} and x > 1; find :

(i) x−1xx - \dfrac{1}{x}

(ii) x3−1x3x^3 - \dfrac{1}{x^3}

Expansions

23 Likes

Answer

(i) Given,

⇒x2+1x=313⇒x+1x=103.\Rightarrow \dfrac{x^2 + 1}{x} = 3\dfrac{1}{3} \\[1em] \Rightarrow x + \dfrac{1}{x} = \dfrac{10}{3}.

By formula,

⇒(x+1x)2−(x−1x)2=4⇒(103)2−(x−1x)2=4⇒1009−(x−1x)2=4⇒(x−1x)2=1009−4⇒(x−1x)2=100−369⇒(x−1x)2=649⇒x−1x=649⇒x−1x=83=223.\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \Big(\dfrac{10}{3}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \dfrac{100}{9} - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = \dfrac{100}{9} - 4 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = \dfrac{100 - 36}{9} \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = \dfrac{64}{9} \\[1em] \Rightarrow x - \dfrac{1}{x} = \sqrt{\dfrac{64}{9}} \\[1em] \Rightarrow x - \dfrac{1}{x} = \dfrac{8}{3} = 2\dfrac{2}{3}.

Hence, x−1x=223x - \dfrac{1}{x} = 2\dfrac{2}{3}.

(ii) By formula,

⇒x3−1x3=(x−1x)3+3(x−1x)=(83)3+3×83=51227+243=512+21627=72827=262627.\Rightarrow x^3 - \dfrac{1}{x^3} = \Big(x - \dfrac{1}{x}\Big)^3 + 3\Big(x - \dfrac{1}{x}\Big) \\[1em] = \Big( \dfrac{8}{3}\Big)^3 + 3 \times \dfrac{8}{3} \\[1em] = \dfrac{512}{27} + \dfrac{24}{3} \\[1em] = \dfrac{512 + 216}{27} \\[1em] = \dfrac{728}{27} \\[1em] = 26\dfrac{26}{27}.

Hence, x3−1x3=262627x^3 - \dfrac{1}{x^3} = 26\dfrac{26}{27}.

Answered By

15 Likes

Related Questions