Tricky

Algebra Level 3

If ( a + 1 a ) 2 = 3 ( a + \frac{1}{a} )^2 = 3 , then a 3 + 1 a 3 = a^3 + \frac{1}{ a^3 } =

This problem is a part of the set - 1's & 2's & QuEsTiOnS
0 0 and 6 3 6 \sqrt{3} 6 3 6 \sqrt{3} 3 3 3 \sqrt{3} none of these

Sakanksha Deo by Sakanksha Deo , 23, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Sravanth C.
Mar 23, 2015

According to the question, we cay say that, a + 1 a = 3 a+\frac { 1 }{ a } =\sqrt { 3 }

Or, ( a + 1 a ) 3 = 3 3 { \left( a+\frac { 1 }{ a } \right) }^{ 3 }=3\sqrt { 3 }

Or, 3 3 = a 3 + ( 1 a ) 3 + 3 × a × 1 a ( a + 1 a ) 3\sqrt { 3 } \quad =\quad { a }^{ 3 }+{ \left( \frac { 1 }{ a } \right) }^{ 3 }+3\times a\times \frac { 1 }{ a } \left( a+\frac { 1 }{ a } \right)

Or, 3 3 = a 3 + ( 1 a ) 3 + 3 3 3\sqrt { 3 } \quad =\quad { a }^{ 3 }+{ \left( \frac { 1 }{ a } \right) }^{ 3 }+3\sqrt { 3 }

Or, a 3 + ( 1 a ) 3 = 0 { a }^{ 3 }+{ \left( \frac { 1 }{ a } \right) }^{ 3 }=\quad 0

1 Helpful 0 Interesting 0 Brilliant 0 Confused

This would be relevant.

Omkar Kulkarni - 6 years, 2 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...