Question
If $${x^2} + \frac{1}{{{x^2}}} = 1{\text{,}}$$ then the value of $${x^{102}}$$ $$ + $$ $${x^{96}}$$ $$ + $$ $${x^{90}}$$ $$ + $$ $${x^{84}}$$ $$ + $$ $${x^{78}}$$ $$ + $$ $${x^{72}}$$ $$ + $$ $$5$$ is?
Answer: Option B
$$\eqalign{
& {x^2} + \frac{1}{{{x^2}}} = 1 \cr
& {\text{Then, }}{\left( {x + \frac{1}{x}} \right)^2} = 1 + 2 \cr
& \Rightarrow x + \frac{1}{x} = \sqrt 3 \cr
& \Rightarrow {x^3} + \frac{1}{{{x^3}}} \cr
& = {\left( {\sqrt 3 } \right)^3} - 3\sqrt 3 \cr
& = 3\sqrt 3 - 3\sqrt 3 \cr
& = 0 \cr
& {\text{Then,}} \cr
& {x^{102}} + {x^{96}} + {x^{90}} + {x^{84}} + {x^{78}} + {x^{72}} + 5 \cr} $$
$$ = {x^{96}}\left( {{x^6} + 1} \right) + $$ $${x^{84}}\left( {{x^6} + 1} \right) + $$ $${x^{72}}\left( {{x^6} + 1} \right) + $$ $$5$$
$$ = 5$$
Was this answer helpful ?
$$\eqalign{
& {x^2} + \frac{1}{{{x^2}}} = 1 \cr
& {\text{Then, }}{\left( {x + \frac{1}{x}} \right)^2} = 1 + 2 \cr
& \Rightarrow x + \frac{1}{x} = \sqrt 3 \cr
& \Rightarrow {x^3} + \frac{1}{{{x^3}}} \cr
& = {\left( {\sqrt 3 } \right)^3} - 3\sqrt 3 \cr
& = 3\sqrt 3 - 3\sqrt 3 \cr
& = 0 \cr
& {\text{Then,}} \cr
& {x^{102}} + {x^{96}} + {x^{90}} + {x^{84}} + {x^{78}} + {x^{72}} + 5 \cr} $$
$$ = {x^{96}}\left( {{x^6} + 1} \right) + $$ $${x^{84}}\left( {{x^6} + 1} \right) + $$ $${x^{72}}\left( {{x^6} + 1} \right) + $$ $$5$$
$$ = 5$$
Was this answer helpful ?
More Questions on This Topic :
Question 4. Simplified value of **Hidden Equation** = ?
Question 5. If **Hidden Equation**
Question 7. If **Hidden Equation** ....
Question 10. If **Hidden Equation** = ?....
Submit Solution