algebra!

Algebra Level 3

What are the answers to these questions?

16, 6627 18, 6621 12, 6409 14, 6407

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1 solution

Sanjeet Raria
Aug 27, 2014

x 2 + 1 = 3 x x^2+1=3x => ( x + 1 x ) = 3 (x+\frac{1}{x})=3 .Now ( x + 1 x ) 3 = x 3 + 1 x 3 + 3 ( x + 1 x ) (x+\frac{1}{x})^3=x^3+\frac{1}{x^3}+3(x+\frac{1}{x}) putting the values x 3 + 1 x 3 = 18 x^3+\frac{1}{x^3}=\boxed{18} ..... The option having this value, as you can see, is unique. So no need to solve that bulky expression.

Please....please....tell us how to solve that too :P

Krishna Ar - 6 years, 9 months ago

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x 7 + 1 ( x 7 ) = ( x 3 + 1 ( x 3 ) ) ( x 4 + 1 ( x 4 ) = x 7 + x + 1 x + 1 ( x 7 ) x^{7} + \frac{1}{(x^{7})} = (x^{3} + \frac{1}{(x^{3})})(x^{4} + \frac{1}{(x^{4})} = x^{7} + x + \frac{1}{x} + \frac{1}{(x^{7})}

x 9 + 1 ( x 9 ) x^{9} + \frac{1}{(x^{9})} = [ x 3 + 1 ( x 3 ) ] 3 3 ( 1 ( x 3 ) + x 3 ) [x^{3} + \frac{1}{(x^{3})}]^{3} - 3( \frac{1}{(x^{3})} + x^{3})

now just substitute the values you will get your answer

U Z - 6 years, 8 months ago

Sorry i could not respond you due to some problem. Thanks to @Megh. Just take a look at my note You can find the clue to deal with such expression.

Try this also.

Sanjeet Raria - 6 years, 8 months ago

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