For-two-non-negative-real-numbers-a-6-b-6-2-Find-a-b-such-that-3-a-b-1-5-ab-

Question Number 155377 by mathdanisur last updated on 29/Sep/21

For two non - negative real numbers :

a^{6} + b^{6} = 2

Find a; b such that 3 (a + b) = 1 + 5 \sqrt{ab}

Answered by MJS_new last updated on 30/Sep/21

the only real solution is a = b = 1

Commented by mathdanisur last updated on 30/Sep/21

Thankyou S er, how if possible

Answered by MJS_new last updated on 30/Sep/21

3 (a + b) = 1 + 5 \sqrt{a b}

let a = u - v \land b = u + v

6 u = 1 + 5 \sqrt{u^{2} - v^{2}} \Rightarrow v^{2} = - \frac{11 u^{2} - 12 u + 1}{25}

a^{6} + b^{6} - 2 = 0

u^{6} + 15 u^{4} v^{2} + 15 u^{2} v^{4} + v^{6} - 1 = 0

inserting & transforming

u^{6} - \frac{279}{679} u^{5} - \frac{2985}{2716} u^{4} + \frac{405}{2716} u^{3} + \frac{45}{21728} u^{2} - \frac{9}{10864} u + \frac{7813}{21728} = 0

{(u - 1)}^{2} (u^{4} + \frac{1079}{679} u^{3} + \frac{2931}{2716} u^{2} + \frac{1951}{2716} u + \frac{7813}{21728}) = 0

this has no other real solution than

u = 1

\Rightarrow

v = 0

\Rightarrow

a = b = 1

Commented by mathdanisur last updated on 30/Sep/21

Very impressive S er thank you