I need to show that there exists no non zero real numbers a,b such that the SqRt of a^2 plus b^2 is equal to the CubeRt of a^3 plus b^3... I have been pluggin in everything cant seem to find the right combination !

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- Apr 16th 2009, 01:21 PMnikie1o2Existence proofs
I need to show that there exists no non zero real numbers a,b such that the SqRt of a^2 plus b^2 is equal to the CubeRt of a^3 plus b^3... I have been pluggin in everything cant seem to find the right combination !

- Apr 16th 2009, 01:36 PMMoo
Hello,

$\displaystyle (\sqrt{a^2+b^2})^6=(\sqrt[3]{a^3+b^3})^6$

$\displaystyle (a^2+b^2)^3=(a^3+b^3)^2$

$\displaystyle a^6+b^6+3a^2b^4+3a^4b^2=a^6+b^6+2a^3b^3$

$\displaystyle 3a^2b^4+3a^4b^2-2a^3b^3=0$

$\displaystyle a^2b^2(3b^2-2ab+3a^2)=0$

since a and b are not equal to 0, this is equivalent to :

$\displaystyle 3b^2-2ab+3a^2=0$

$\displaystyle (\sqrt{3} b)^2-2\cdot \frac{a}{\sqrt{3}} \cdot (\sqrt{3} b)+3a^2=0$

$\displaystyle \left(\sqrt{3} b-\frac{a}{\sqrt{3}}\right)^2-\frac{a^2}{3}+3a^2=0$

$\displaystyle \left(\sqrt{3} b-\frac{a}{\sqrt{3}}\right)^2+\frac{10}{3} \cdot a^2=0$

this is true if and only if $\displaystyle \sqrt{3} b-\frac{a}{\sqrt{3}}=0$ and $\displaystyle a=0$, because both terms in the sum are positive.

And this is not possible since $\displaystyle a \neq 0$