Manipulation

Algebra Level 3

Let $a$ and $b$ with $a>b>0$ be real numbers satisfying $a^3+b^3=a+b$ . Find $\sqrt{\dfrac{a}{b}+\dfrac{b}{a}-\dfrac{1}{ab}}$ .

Give your answer to 3 decimal

This question is part of the set All-Zebra

2 solutions

A Former Brilliant Member
May 3, 2016

$\Rightarrow a^3+b^3=a+b$

$(a+b)(a^2+b^2-ab)=a+b$

$a^2+b^2-ab=1$

$a^2+b^2-1=ab$

We need to find: $\sqrt{\dfrac{a}{b}+\dfrac{b}{a}-\dfrac{1}{ab}}=\sqrt{\dfrac{a^2+b^2-1}{ab}}$

$\sqrt{\dfrac{ab}{ab}}=\boxed{1}$

Did the same way.

Niranjan Khanderia - 5 years, 1 month ago

This looks good. :). Nice solution.

Abhay Tiwari - 5 years, 1 month ago

Abhay Tiwari
May 3, 2016

$(a+b)^{3}=a^3+b^3+3ab(a+b)$

$(a+b)^{3}=(a+b)+3ab(a+b)$

$(a+b)^{2}=(1+3ab)$

$a^2+b^2+2ab=1+3ab$

$a^2+b^2=1+ab$

Dividing $L.H.S.$ and $R.H.S.$ by $ab$ .

$\dfrac{a}{b}+\dfrac{b}{a}=\dfrac{1}{ab}+1$

$\dfrac{a}{b}+\dfrac{b}{a}-\dfrac{1}{ab}=1$

$\sqrt{\dfrac{a}{b}+\dfrac{b}{a}-\dfrac{1}{ab}}=\boxed{1.000}$