A fascinating equation!!

Algebra Level 2

If ( a 2 + b 2 ) 3 = ( a 3 + b 3 ) 2 (a^{2} + b^{2})^{3} = (a^{3} + b^{3})^{2} where a and b are real numbers other than zero, then find the numerical or absolute value of 3 ( a b + b a ) 3( \frac{a}{b} + \frac{b}{a}) .


The answer is 2.000.

Ninad Akolekar by Ninad Akolekar , 23, India

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3 solutions

Sandeep Rathod
Nov 16, 2014

a 6 + b 6 + 3 a 2 b 2 ( a 2 + b 2 ) = a 6 + b 6 + 2 a 3 b 3 a^{6} + b^{6} + 3a^{2}b^{2}(a^{2} + b^{2}) = a^{6} + b^{6} + 2a^{3}b^{3}

a 2 b 2 ( 3 a 2 + 3 b 2 − 2 a b ) = 0 a^{2}b^{2}(3a^{2} + 3b^{2} - 2ab) = 0

a 3 b 3 ( 3 a b + 3 b a − 2 ) = 0 a^{3}b^{3}(3\frac{a}{b} + 3\frac{b}{a} - 2) = 0

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Expanding the equation,we get: a 6 + 3 a 4 b 2 + 3 a 2 b 4 + b 6 = a 6 + 2 a 3 b 3 + b 6 a^6+3a^4b^2+3a^2b^4+b^6=a^6+2a^3b^3+b^6 Simplifying,we get: 3 a 2 b 2 ( a 2 + b 2 ) = 2 a 3 b 3 3a^2b^2(a^2+b^2)=2a^3b^3 3 ( a 2 + b 2 ) = 2 a b 3(a^2+b^2)=2ab 3 ( a 2 + b 2 a b ) = 2 3\left(\frac{a^2+b^2}{ab}\right)=2 3 ( a 2 a b + b 2 a b ) = 2 3\left(\frac{a^2}{ab}+\frac{b^2}{ab}\right)=2 3 ( a b + b a ) = 2 3\left(\frac{a}{b}+\frac{b}{a}\right)=\boxed{2}

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Harsh Soni
Nov 16, 2014

Soooo easy.....

0 Helpful 0 Interesting 0 Brilliant 0 Confused

This is a solution?

Abdur Rehman Zahid - 6 years, 6 months ago

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