2022/008) Given $\sin^3 x + \cos ^3 x= \frac{1}{\sqrt{2}}$ find $\sin \, 2x$

 we have  $\sin^3 x + \cos ^3 x= \frac{1}{\sqrt{2}}$

letting $\sqrt{2} \sin\,x = s$ and $\sqrt{2} \cos\, x = c$

we have $s^3 + c^3 = 2 \sqrt{2} \frac{1}{\sqrt{2}}= 2$

or $s^3 + c^3 = 2\cdots(1)$

and $s^2 + c^2 = 2\cdots(2)$

we are interested to find $\sin\,2x$ or sc

squaring (1) we get $s^6 + c^6 + 2s^3c^3 = 4$

and cubing (2) we get $s^6 + c^6 + 3s^2c^2(s^2 + c^2) = 8$

or $s^6 + c^6 + 6s^2c^2 = 8$

from (3) and (4) we have letting sc = x

$2x^3 - 6x^2 + 4 = 0$

or $x^3 - 3x^2 + 2$   this gives x = 1

and deviding by x-1 for factoring we get $(x-1)(x^2 - 2x -2) = 0$

so x = 1 or $1\pm \sqrt{3}$ but $1 + \sqrt{3}$ being above 1 is not admissible

so x = 1 or $1-\sqrt{3}$