SSC CGL 24 Aug 2021 Shift 3 Paper

Question : 53 of 100

Marks: +1, -0

Given that,

x^{8}-34 x^{4}+1=0, x > 0

. What is the value of

(x^{3}-x^{-3})

Solution:

Given,

x^{8}-34 x^{4}+1=0

x^{8}+1=34 x^{4}

Divided by

x^{4}

on both sides,

\;x^{4}+\;{1}/{x^{4}}=34

⇒\;x^{4}+\;{1}/{x^{4}}+2=34+2

⇒(x^{2}+\;{1}/{x^{2}})^{2} \;\; =36

⇒ (x^{2}+\;{1}/{x^{2}})=√{36}=6

⇒ (x^{2}+\;{1}/{x^{2}})=6

\;\text " Now, "\;(x-\;{1}/{x})^{2}=x^{2}+\;{1}/{x^{2}}-2 × x × \;{1}/{x}

(x-\;{1}/{x})^{2}=6-2=4

(x-\;{1}/{x})=√{4}=2

Take cube on both sides,

(x-\;{1}/{x})^{3}=(2)^{3}

∵ \;\; x^{3}-\;{1}/{x^{3}}-3 × x × \;{1}/{x}(x-\;{1}/{x})=8

\;\; [ ∵ (a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)]

⇒ \;\; x^{3}-\;{1}/{x^{3}}=8+3 × 2=8+6

⇒ \;\; x^{3}-\;{1}/{x^{3}}=14

x^{3}-x^{-3}=14

Go to Question:

Prev Question

Next Question