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Question Number 157790 by MathSh last updated on 28/Oct/21

	Answered by MJS_new last updated on 28/Oct/21
	$let c=cos x ((32c^5 −40c^3 +9c))^(1/3) +((16c^5 −20c^3 +4c))^(1/3) =(c)^(1/3) u+v=w ∣^3 u^3 +3u^2 v+3uv^2 +v^3 =w^3 u^3 +3uv(u+v)+v^3 =w^3 u^3 +3uvw+v^3 =w^3 3uvw=w^3 −(u^3 +v^3 ) 27u^3 v^3 w^3 =(w^3 −(u^3 +v^3 ))^3 inserting we get 108c^3 (4c^4 −5c^2 +1)(32c^4 −40c^2 +9)=−1728c^3 (4c^4 −5c^2 +1)^3 ⇒ c=0∨4c^4 −5c^2 +1=0 32c^4 −40c^2 +9=−16(4c^4 −5c^2 +1)^2 let c^2 =d+(5/8) 32d^2 −(7/2)=−(((64d^2 −9)^2 )/(16)) d^4 −(5/(32))d^2 +((25)/(4096))=0 (d^2 −(5/(64)))^2 =0 d=±((√5)/8) c^2 =((5±(√5))/8) now we have c=0 c=±1 c=±(1/2) c=±(√((5−(√5))/8)) c=±(√((5+(√5))/8)) testing leads to c=0∨c=±(1/2)∨c=±1 for 0≤x<2π we get x∈{0, 2, 3, 4, 6, 8, 9, 10}×(π/6)$
	$let c = \cos x$ $\sqrt[3]{32 c^{5} - 40 c^{3} + 9 c} + \sqrt[3]{16 c^{5} - 20 c^{3} + 4 c} = \sqrt[3]{c}$ $u + v = w ∣^{3}$ $u^{3} + 3 u^{2} v + 3 {u v}^{2} + v^{3} = w^{3}$ $u^{3} + 3 u v (u + v) + v^{3} = w^{3}$ $u^{3} + 3 u v w + v^{3} = w^{3}$ $3 u v w = w^{3} - (u^{3} + v^{3})$ $27 u^{3} v^{3} w^{3} = {(w^{3} - (u^{3} + v^{3}))}^{3}$ $inserting we get$ $108 c^{3} (4 c^{4} - 5 c^{2} + 1) (32 c^{4} - 40 c^{2} + 9) = - 1728 c^{3} {(4 c^{4} - 5 c^{2} + 1)}^{3}$ $\Rightarrow c = 0 \lor 4 c^{4} - 5 c^{2} + 1 = 0$ $32 c^{4} - 40 c^{2} + 9 = - 16 {(4 c^{4} - 5 c^{2} + 1)}^{2}$ $let c^{2} = d + \frac{5}{8}$ $32 d^{2} - \frac{7}{2} = - \frac{{(64 d^{2} - 9)}^{2}}{16}$ $d^{4} - \frac{5}{32} d^{2} + \frac{25}{4096} = 0$ ${(d^{2} - \frac{5}{64})}^{2} = 0$ $d = \pm \frac{\sqrt{5}}{8}$ $c^{2} = \frac{5 \pm \sqrt{5}}{8}$ $now we have$ $c = 0$ $c = \pm 1$ $c = \pm \frac{1}{2}$ $c = \pm \sqrt{\frac{5 - \sqrt{5}}{8}}$ $c = \pm \sqrt{\frac{5 + \sqrt{5}}{8}}$ $testing leads to$ $c = 0 \lor c = \pm \frac{1}{2} \lor c = \pm 1$ $for 0 ⩽ x < 2 π we get$ $x \in {0, 2, 3, 4, 6, 8, 9, 10} \times \frac{π}{6}$
	Commented by MathSh last updated on 28/Oct/21

	$Perfect dear Ser, thank you very much$
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