[tex]\cos\dfrac\pi3=\dfrac12[/tex]
so you have
[tex]\dfrac3{5\cos^3\frac\pi3}=\dfrac3{5\left(\frac12\right)^3}=\dfrac3{\frac58}=\dfrac{24}5[/tex]
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If you don't remember the value of [tex]\dfrac\pi3[/tex] off the top of your head, it's possible to derive it with some identities and knowing that [tex]\cos\pi=-1[/tex].
Consider the expression [tex]\cos3x[/tex]. With the angle sum identity, we have
[tex]\cos3x=\cos x\cos2x-\sin x\sin2x[/tex]
and the double angle identities give
[tex]\cos3x=\cos x(\cos^2x-\sin^2x)-2\sin^2x\cos x[/tex]
Write everything in terms of cosine:
[tex]\cos3x=\cos x(2\cos^2x-1)-2(1-\cos^2x)\cos x[/tex]
[tex]\cos3x=4\cos^3x-3\cos x[/tex]
Now let [tex]x=\dfrac\pi3[/tex]. Then
[tex]\cos\pi=4\cos^3\dfrac\pi3-3\cos\dfrac\pi3[/tex]
Let [tex]y=\cos\dfrac\pi3[/tex]. Then
[tex]-1=4y^3-3y[/tex]
[tex]4y^3-3y+1=0[/tex]
The rational root theorem suggests some possible roots are
[tex]\pm\dfrac14,\pm\dfrac12,\pm1[/tex]
and checking all of these, we find that [tex]y=\dfrac12[/tex] is among the solution set. In fact,
[tex]4y^3-3y+1=(y+1)\left(y-\dfrac12\right)^2=0\implies y=-1\text{ or }y=\dfrac12[/tex]
We have [tex]\cos x=-1[/tex] only for odd multiples of [tex]\pi[/tex], so it follows that
[tex]\cos\dfrac\pi3=\dfrac12[/tex]
